I think I can figure out the leading digit or the units digit, but how would one find the tens digit of, say, $$7^{100}$$ without having a calculator or web app that actually displays the full integer.
Is there a way to find the tens digit given a scientific calculator that can only display the answer in scientific notation?
Also, I have no idea what to tag this question.
The tens digit of $n$ can be read off of $n\pmod {100}$. In this case, $7^4\equiv 1 \pmod {100}$ so $7^{100}\equiv 1 \pmod {100}$, so the answer is $0$.
$7^{100}$ turns out to be a bit boring, so let's work up a method for finding the last two digits of $13^{37}$ with your calculator. Everyone else has given results using modular arithmetic, so I'll just make a basic explanation/exploration of what is possible.
$13^2 = 169$. Whatever we multiply this by, the $100$ portion is not going to affect the last two digits, so we can throw that away in any further calculation and just focus on the $69$. This is true for every calculation, so we could just repetitively multiple by $13$, subtract off the hundreds, do it again:
$$13\cdot 13 = 169 \to 69\cdot 13 = 897 \to 97\cdot 13 = 1261 \to 61\cdot 13 = 793\\ \to 93\cdot 13 = 1209 \to 9\cdot 13 = 117 \to 17\cdot 13 = 221 \to 21\cdot 13 = 273 \to 73$$
So here we've got as far as finding that the last two digits of $13^9$ are $73$.
We don't have to go one step at a time, though. The last two digits of $13^6$ were $09$, which would mean that the last two digits of $13^{12}$ will be $09^2=81$. Let's just restart the process above to check that:
$$73\cdot 13 = 949 \to 49\cdot 13 = 637 \to 37\cdot 13 = 481 \to 81$$
Yes. Similarly we can get the last two digits of $13^{18}$ from $09\cdot 81 = 729 \to 29$, and the last two digits of $13^{36}$ from $29^2 = 841\to 41$. Then the last two digits of $13^{37}$ can be found with $41\cdot 13 = 533\to 33$.
With $7^{100}$, which I said was boring, the corresponding first few steps gives us
$$7.\cdot 7 = 49 \to 49.\cdot 7 = 343 \to 43\cdot 7 = 301 \to 01 $$
and since the last two digits of $7^4$ are $01$, we can immediately say that $01^{25}=01$ and thus the tens digit of $7^{100}$ is zero.
Suppose we have a number in the decimal number system with the following representation: $a_na_{n-1}\ldots a_1a_0$, where all the $a_i \in \{0,1, \ldots, 9\}$. This means that the number is $$a_n\cdot 10^n + a_{n-1}\cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0\cdot 10^0$$ so if you want to find the digit $a_1$, then you have to reduce the above number modulo 100 (since this will leave you with $a_1 \cdot 10 + a_0$).
Looking at your number, we want to compute $7^{100} \mod 100$. Because $7^4 = 2401$, we have that $7^4 \equiv 1 \mod 100$ and hence we find that $$7^{100} \equiv 7^{4 \cdot 25} \equiv (7^4)^{25} \equiv 1^{25} \equiv 1 \mod 100.$$ So that $a_1\cdot 10 + a_0 = 1$ and hence $a_1 = 0, a_0 = 1$. So the second last digit is $0$.
Now let me ask you a question: if I were to ask you how to find the '100' digit, how would you do this?
To find unit digit we use mod 10.
To find tens digit mod 100 is used.
$7^4 \equiv 1 \pmod {100}$
$7^{100} \pmod{100} \equiv (7^4)^{25} \pmod {100} \equiv 1^{25} \pmod {100} \equiv 0$
So answer is zero.
