a) Let $a$ be an odd integer. If $2^a \equiv 1\ (\text{mod}\ n)$, then $ \varphi(n)|a$.
b) Let $p$ be an odd prime. There exists $a \in \mathbb{Z} >0$ with $\varphi(a) = 2p$.
I think a is false and b is true, but I'm not sure. I'd appreciate any help, thanks!
Yes, the first one is indeed false. For instance, $2^9\equiv1\pmod{7}$, but $\varphi(7)=6$ and $6\nmid9$.
And you were right about the other one too.
I'd try to think why something might be true. And if I end up with an insufficient reason, try to exploit that to construct a counter example.
1) Why would $2^a\equiv 1\pmod n \implies \phi(n)|a$? Well, if $\gcd(2,n)=1$ the Euler's theorem says $2^{\phi(n)} \equiv 1 \pmod n$. And if $\gcd(2,n)\ne 1$ then $2^a\equiv 1 \pmod n$ is impossible.
If $\phi(n)|a$ then we definitely have $2^a\equiv 1\pmod n$ but not necessarily the other way around. But we do have if $2^a\equiv 1\pmod n$ that means $a$ is a multiple of the order of $2$ and the order of $2$ will divide $\phi(n)$.
So if we can find an $n$ were the order of $2$ via modulo $n$ multiplication is less than $\phi(n)$ we will be done.
In other words, to find a counter example we just need an $n$ where $2$ is not a primitive root in $\mathbb Z_n$. $2^3\equiv 1 \pmod 7$ and indeed $2^k\equiv 1\pmod(2^k-1)$ and and $3 < 7-1$ and $k < 2^k-2$ if $k>2$, then $2$ is not a primitive root $\mod 7$ or of any $2^k-1; k>2$. And then $3|\phi(7)=6$ we do not have $\phi(7)=6|3$. (Nor can $\phi(2^k-1)|k$ if $k> 2$. Apparently.)
... Okay that was a lot of talk for a very simple counter-example....
2) Why would $\phi(n) = 2p$? Well, $\phi(n)$ is multiplicative and if $a = \prod p_i^{a_i}$ is the prime factorization of $n$ then $\phi(n) = [\prod (p_i-1)]\prod p_i^{a_i -1}$.
So can $[\prod (p_i-1)]\prod p_i^{a_i -1}= 2p$?
Note: If $m$ of the $p_k$s are odd then $p_k-1$ is even and $2^m|\phi(n)$. So as $\phi(n) = 2p$ we have at most one odd prime factor of $n$.
So either $n = 2^k$ or $n = 2^kq^j$ for some integers $k,j$ and odd prime $q$.
$\phi(2^k) = 2^{k-1}\ne 2p$ so that's out.
And $\phi (2^kq^j)= 2^{k-1}(q-1)q^{j-1}=2p$. $q-1$ is even so $k=0$ or $1$ and if $p > 3$ then $q-1$ has a prime factor which must be $p$.
So if $p > 3$ and $\phi(n) = 2p$ then there is a prime $q$ where $q= 2p + 1$. So to find a counterexample we can just find an odd prime where $2p + 1$ is composite. For example $p =7$.
There is no $n$ where $\phi(n) = 14$.
