Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$ and show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module.

1) I think you made this much harder than it has to be. $\mathbb{Q}$ is a divisible $\mathbb{Z}$-module. This implies that if $f:\mathbb{Q}\to\mathbb{Z}$, then $f(\mathbb{Q})$ is a divisible $\mathbb{Z}$-submodule of $\mathbb{Z}$. How many of those do you know?

2) Hint: PIDs have the nice property that a submodule of a free module is free. With this, what can you say about projective $\mathbb{Z}$-modules? Namely note that they are all (BLANK). After you've done this, explain why $\mathbb{Q}$-can't be (BLANK) by 1) and thus can't be projective.

2) If $\mathbb{Q}$ is a projective $\mathbb{Z}$-module, then it is direct summand of a free module, which is isomorphic to $\mathbb{Z}^{(I)}:=\oplus_{i\in I}\mathbb{Z}$; for some set $I$.

In this case there is a injective homomorphism $\iota:\mathbb{Q}\hookrightarrow \mathbb{Z}^{(I)}$. But note that, for each $i\in I$, if $\pi_i:\mathbb{Z}^{(I)}\to\mathbb{Z}$ is the $i$-ith projection, $\pi_i\iota:\mathbb{Q}\to\mathbb{Z}$ is a homomorphism and, therefore it is the null homomorphism.

Thus, for any $x\in \mathbb{Q}$, $\pi_{i}\iota(x)=0$ for every $i\in I$. Then $\iota(x)=0$. Therefore $\iota$ is also the null map - which contradicts the hypothesis that it is a injective homomorphism.