For the first question the answer is yes, and an upper bound is 16.
We define a nearly good number to be such that the last digit is not the last digit
of any of its proper divisors. We have the following result:
$$
\text{Every nearly good number is the product of at most
4 prime numbers.}\ \ \ \ \ (*)
$$
Hence every nearly good number (and also every good number) has at most 16 divisors.
The proof of (*) is very long, but straightforward. Here a sketch of the proof:
Assume that $n>10$ is a nearly good number. Write $\bar n$ for the last digit (or, by abuse of notation, also for the class in $\Bbb{Z}/10\Bbb{Z}$).
We first prove that if $\bar n=9$, then it has only two prime factors at most.
In fact, the only valid factorizations of $\bar 9$ in two factors are
$$
\bar9=\bar 3\cdot \bar 3\quad\text{and}\quad \bar9=\bar 7\cdot \bar 7
$$
Since the only valid factorization of $\bar 3$ is $\bar 3=\bar 9\cdot \bar 7$ and the only
valid factorization of $\bar 7$ is $\bar 7=\bar 9\cdot \bar 3$, both factors must be prime factors.
Arguing similarly, one obtains that if $\bar n=3$, then it has at most three prime factors, and in that case
$\bar n=\bar 7\cdot\bar 7\cdot\bar 7$. And it is also straightforward to show that if $\bar n=7$, then it has
at most three prime factors, and in that case
$\bar n=\bar 3\cdot\bar 3\cdot\bar 3$.
Now comes the clumsy part:
If $\bar n=8$, then the only valid factorizations in two factors are
$$
\bar n=\bar 2\cdot \bar 4,\quad\bar n=\bar 3\cdot \bar 6,\quad\bar n=\bar 2\cdot \bar 9 \quad\text{and}\quad \bar n=\bar 4\cdot \bar 7.
$$
In the four cases one shows that there are only three prime factors possible. For example, if we assume $\bar 8=\bar 2\cdot \bar 4$, then the factor $\bar 2$ correspond to the prime factor 2 (Else the factor is even and of the form $\bar m\cdot 2$ with $m>1$, hence $\bar 8=\bar m\cdot \bar 2\cdot\bar 4$ and the proper factor
$\bar 2\cdot \bar 4$ has 8 as last digit).
Since the factor with $\bar 4$ is even, it has a factorization $\bar 4=\bar m\cdot \bar 2$, and once again $\bar 2$ is the prime factor 2, and $\bar m=2$ or $\bar m=7$. In the first case all three factors are 2 and in the second case
we cannot factorize $\bar 7$ (else $\bar 7=\bar 3\cdot \bar 9$, and the original number has a proper factor corresponding to $\bar 2\cdot \bar 9$, with last digit 8). The other cases are similar and prove that if $\bar n=8$, then there are at most three factors.
Finally one deals with the cases $\bar n=4$ and $\bar n=6$. Here
the valid factorizations for $\bar n=4$ in two factors are
$$
\bar n=\bar 2\cdot \bar 2=\bar 7\cdot \bar 2=\bar 3\cdot \bar 8=\bar 9\cdot \bar 6.
$$
One verifies that there are at most 4 prime factors, with the only possible configuration
$$
\bar n=\bar 3\cdot \bar 3\cdot\bar 3\cdot \bar 2.
$$
Similarly
the valid factorizations for $\bar n=6$ in two factors are
$$
\bar n=\bar 2\cdot \bar 3=\bar 8\cdot \bar 2=\bar 4\cdot \bar 4=\bar 9\cdot \bar 4=\bar 7\cdot \bar 8.
$$
One verifies that there are at most 4 prime factors, and the only possible configurations are then
$$
\bar n=\bar 2\cdot \bar 2\cdot \bar 2\cdot \bar 7,\quad
\bar n=\bar 2\cdot \bar 2\cdot \bar 7\cdot \bar 7\quad
\text{and}\quad \bar n=\bar 2\cdot \bar 7\cdot\bar 7\cdot \bar 7.
$$
So we have proved: If $n$ is a (nearly) good number, then $d(n)\le k(\bar n)$ where
$$
k(9)=4,\quad k(3)=8,\quad k(7)=8,\quad k(8)=8,\quad k(4)=16,\quad\text{and}\quad k(6)=16.
$$
Clearly $\bar n=0,1,2,5$ is not possible if $n>10$.
Although there are good numbers with four prime factors, I have never seen any with more than 8 divisors. It is highly probable that the maximum $d(n)$ for good numbers is 8. It is nearly impossible that it is 16, and there is a small chance that it is 9 or 12: If $d(n)>8$, then $n$ must have 4 prime factors and the valid configurations are
$$
\bar n=\bar 2\cdot \bar 3\cdot \bar 3\cdot \bar 3,\quad
\bar n=\bar 2\cdot \bar 7\cdot \bar 7\cdot \bar 7\quad
\text{and}\quad \bar n=\bar 2\cdot \bar 2\cdot\bar 7\cdot \bar 7.
$$
In the first case $1,2,3,6,7,8,9\ne Dig(n)$, and the combinations $50$ and $40$ in the string of digits of $n$ are impossible, hence the decimal representation of $n$ is a string with only 5's and 4's. How many of these numbers are product of 4 primes ending in 3,3,3,2? At least 54 is one of them, but as I said, I find it nearly impossible that there is a bigger good number like that.
In the second case $1,2,3,4,7,8,9\ne Dig(n)$, and the combinations $50$ and $60$ in the string of digits of $n$ are impossible, hence the decimal representation of $n$ is a string with only 5's and 6's. How many of these numbers are product of 4 primes ending in 7,7,7,2? I find it nearly impossible that there is number like that.
In the third case $1,2,4,7,8,9\ne Dig(n)$, hence the decimal representation of $n$ is a string with only 0's,3's,5's and 6's. If the two primes ending in 7 are equal, then $n$ has 9 divisors, else it has 12. Although the chance seems bigger than in the two previous cases, the bigger the number are that one tests, the more digits can yield a failure in being a good number, so the chance is very small.
$\textbf{Second question:}$ For the second question the answer is yes, and $56,57,58,59$ is the biggest such set. In particular, the only sets
of four consecutive good numbers are
$$
\{2,3,4,5\},\{3,4,5,6\},\{4,5,7,8\},\{5,6,7,8\},\{6,7,8,9\},\{56,57,58,59\}.
$$
Note that if $m>0$, then
$10m+5$ and $10m+1$ are not good numbers, so the only possible configuration for four consecutive numbers greater than 10 is
$10m+6,10m+7,10m+8,10m+9$. Moreover, 1,2,3 and 4 are divisors of at least one of them, so the set $Dig(m)$ of the digits of $m$,
cannot contain 1,2,3 or 4, in particular $\bar m\ne 1,2,3,4$. Now we assume $m>0$ and that $10m+6,10m+7,10m+8,10m+9$
are good numbers. We have to prove that $m=5$ is the only possible case.
$\textbf{Remark:}$
a) 3 divides at least one of the four numbers, and it cannot divide $10m+6$, since then $6$ would be a proper divisor of $10m+6$. Hence it divides
$10m+7$ or $10m+8$.
b) If $4|10m+8$, then $\overline{10m+8}=2\cdot 2\cdot \bar 7$, since there are at most three prime factors, and
$10m+8=\bar 2\cdot \bar 2\cdot \bar 2$ leads to $m=0$. In particular, in this case $3\not| 10m+8$.
We will analyse all possible cases of $\bar m$, the last digit of $m$.
$\textbf {CASE}\ \bar m=9$:
Clearly $\bar m\ne 9$, since then $10m+8$ is not a good number:
$$
10m+8=100k+98\quad\text{and}\quad d:=(10m+8)/2=50k+49,
$$
so 9 is the last digit of the proper divisor $d$ and $9\in Dig(10m+8)$, a contradiction.
$\textbf {CASE}\ \bar m=0$:
In this case $4|10m+8=100k+8$, and, since $m>0$, we have that $k$ must be odd (otherwise $8|10m+8$).
Write $k=2j+1$, then $(10m+8)/2=100 j+54$, hence $\bar k\ne 5$ and $(10m+8)/4=50 j+27$, hence $\bar k\ne 7$. Since
$\bar k\ne 1,3$, we have only to discard the case $\bar k=9$. But $3|10m+7$ (by a) and b) above), and so
$\overline{10m+7}= 3\cdot \bar 9$, which implies in this case
$9\in Dig(10m+7)\cap Dig((10m+7)/3)$.
$\textbf {CASE}\ \bar m=8$:
In this case $4|10m+8=100k+88$, and, since $m>0$, we have that $k$ must be odd (otherwise $8|10m+8$).
Write $k=2j+1$, then $(10m+8)/2=100 j+94$, hence $\bar k\ne 9$ and $(10m+8)/4=50 j+47$, hence $\bar k\ne 7$.
Since $\bar k\ne 1,3$, we have only to discard the case $\bar k=5$. But $3|10m+7$ (by a) and b) above), hence,
if $\bar k=5$, we have $10m+7=1000(3r+1)+587$ and so $(10m+7)/3=1000 r+529$ which implies
$5\in Dig(10m+7)\cap Dig((10m+7)/3)$, a contradiction.
$\textbf {CASE}\ \bar m=6$:
In this case $4|10m+8=100k+68$, and, since $m>0$, we have that $k$ must be even (otherwise $8|10m+8$).
Since $\bar k\ne 2,4$, we have to discard the cases $\bar k=0,6,8$. As above $3|10m+7$, hence,
if $\bar k=8$, we have $10m+7=3000r+867$ and so $(10m+7)/3=1000 r+289$ which implies
$8\in Dig(10m+7)\cap Dig((10m+7)/3)$, a contradiction. Similarly,
if $\bar k=0$, we have $10m+7=1000(3r+2)+67$ and so $(10m+7)/3=1000 r+689$ which implies
$6\in Dig(10m+7)\cap Dig((10m+7)/3)$, a contradiction. Finally, if $\bar k=6$, we have
$10m+8=1000r+668$. Then $(10m+8)/2=500r+334$. If $r$ is odd ($r=2j+1$), then
$(10m+8)/2=1000j+834$, impossible ($8\in Dig(10m+8)\cap Dig((10m+8)/2)$), and if
$r$ is even, then $(10m+8)/2=1000j+334$ and so $(10m+8)/4=500j+167$, which leads to the contradiction
$6\in Dig(10m+8)\cap Dig((10m+8)/4)$.
$\textbf {CASE}\ \bar m=7$:
In this case $4|10m+6$.
Write $m=10k+7$. We will discard all possibilities for $\bar k$. Clearly $\bar k\ne 1,2,3,4$.
If $\bar k=9$, then $9\in Dig(10m+8)\cap Dig((10m+8)/2)$, if $\bar k=8$, then
$8\in Dig(10m+6)\cap Dig((10m+6)/2)$ and if $\bar k=7$, then
$8\in Dig(10m+8)\cap Dig((10m+8)/2)$ (note that $(1000j+778)/2=500j+389$).
If $\bar k=6$, we have
$10m+6=1000r+676$. Then $(10m+6)/2=500r+338$. If $r$ is odd ($r=2j+1$), then
$(10m+8)/2=1000j+839$, impossible ($8\in Dig(10m+8)\cap Dig((10m+8)/2)$), and if
$r$ is even, then $(10m+6)/2=1000j+338$ and so $(10m+6)/4=500j+169$, which leads to the contradiction
$6\in Dig(10m+6)\cap Dig((10m+6)/4)$.
If $\bar k=5$ we have two options:
If $3|10m+7$, then we have $10m+7=1000(3r+2)+577$ and so $(10m+7)/3=1000 r+859$ which implies
$5\in Dig(10m+7)\cap Dig((10m+7)/3)$, a contradiction.
If $3|10m+8$, then we have $10m+8=1000(3r+1)+578$ and so $(10m+7)/3=1000 r+526$ which implies
$5\in Dig(10m+8)\cap Dig((10m+8)/3)$, a contradiction.
So we are left with the case $\bar k=0$. Here the only digits that can appear in $m$ are 0,5,6,7:
if $9\in Dig(m)$, then $9\in Dig(10m+8)\cap Dig((10m+8)/2$, since $10m+8=1000r+78$. Similarly,
if $8\in Dig(m)$, then $8\in Dig(10m+6)\cap Dig((10m+6)/2$, since $10m+6=1000r+76$.
Now we discard combinations of two consecutive digits of $m$. Clearly the combination $50$ can be discarded, since
then $m/2$ has the combination 25 or the combination 75, leading to $5\in Dig(10m+6)\cap Dig((10m+6)/2$.
We can also discard $70$: In that case $m/2$ admits the combination 35 (the other possibility $85$ leads to
$8\in Dig(10m+8)\cap Dig((10m+8)/2)$), and so $m/4$ admits the combination 17 or 67, leading to
$7\in Dig(10m+6)\cap Dig((10m+6)/4)$, a contradiction.
The combinations 56 and 76 can be also discarded, since they lead to $8\in Dig(10m+8)\cap Dig((10m+8)/2$.
Resuming the restrictions, we have that on the left of a zero there can be only zero or 6, and to the left of a
6 there can be only a 0 or a 6. Since the two last digits of $m$ are 07, the only digits in the expansion to the left
are 0 and 6. But then $m=600r+7$ and so $3|10m+7$ is impossible and $3|10m+8$ leads to $(10m+8)/3=(6000r+78)/3=2000r+26$,
which implies $6\in Dig(10m+8)\cap Dig((10m+8)/3)$.
$\textbf {CASE}\ \bar m=5$:
In this case $4|10m+6$.
Write $m=10k+5$. We will discard all possibilities for $\bar k$. Clearly $\bar k\ne 1,2,3,4$.
If $\bar k=9$, then $9\in Dig(10m+8)\cap Dig((10m+8)/2)$, if $\bar k=8$, then
$8\in Dig(10m+6)\cap Dig((10m+6)/2)$ and if $\bar k=7$, then
$7\in Dig(10m+8)\cap Dig((10m+8)/2)$ (note that $(1000j+758)/2=500j+379$).
If $\bar k=6$, we have
$10m+6=1000r+656$. Then $(10m+6)/2=500r+328$. If $r$ is odd ($r=2j+1$), then
$(10m+8)/2=1000j+829$, impossible ($8\in Dig(10m+8)\cap Dig((10m+8)/2)$), and if
$r$ is even, then $(10m+6)/2=1000j+328$ and so $(10m+6)/4=500j+164$, which leads to the contradiction
$6\in Dig(10m+6)\cap Dig((10m+6)/4)$.
If $\bar k=5$ we have two options:
If $3|10m+7$, then we have $10m+7=1000(3r+1)+557$ and so $(10m+7)/3=1000 r+519$ which implies
$5\in Dig(10m+7)\cap Dig((10m+7)/3)$, a contradiction.
If $3|10m+8$, then we have $10m+8=3000r+558$ and so $(10m+7)/3=1000 r+186$ which implies
$8\in Dig(10m+8)\cap Dig((10m+8)/3)$, a contradiction.
So we are left with the case $\bar k=0$. Here the only digits that can appear in $m$ are 0,5,6,7:
if $9\in Dig(m)$, then $9\in Dig(10m+8)\cap Dig((10m+8)/2)$, since $10m+8=1000r+58$. Similarly,
if $8\in Dig(m)$, then $8\in Dig(10m+6)\cap Dig((10m+6)/2)$, since $10m+6=1000r+56$.
Now we discard combinations of two consecutive digits of $m$. Clearly the combination $50$ can be discarded, since
then $m/2$ has the combination 25 or the combination 75, leading to $5\in Dig(10m+6)\cap Dig((10m+6)/2)$.
Similarly the combination $70$ can be discarded, since
then $m/2$ has the combination 35 or the combination 85, leading to $5\in Dig(10m+6)\cap Dig((10m+6)/2)$.
The combination $60$ can be discarded, since
then $m/2$ has the combination 30 (the other possibility $80$ leads to
$8\in Dig(10m+8)\cap Dig((10m+8)/2)$), and then $(10m+6)/4$ has the combination $15$ or the combination $65$,
which both imply $5\in Dig(10m+6)\cap Dig((10m+6)/2)$.
Since the two last digits of $m$ are 05, we are left we the only case $m=5$, as desired.
