For any group $G$, you have a bijection between $G$ and the set $Hom(\mathbb{Z},G)$ of group homomorphisms $\mathbb{Z}\to G$.
The bijection is as follows: for $x\in G$, set $h_x:\mathbb{Z}\to $G$, \ m\mapsto x^m$.
Then the desired bijection is $G\to Hom(\mathbb{Z},G), \ x\mapsto h_x.$
First of all, you have to check that $h_x$ is indeed a group homomorphism (easy).
For the injectivity part: if $x,y\in G$ are such that $h_x=h_y$, then $h_x(1)=h_y(1)$, that is $x=y$?
For the surjectivity part: the main idea is that any element of $\mathbb{Z}$ is the sum of several copies of $1$ (or $-1$). Thus, in order to define a homomorphism $\varphi: \mathbb{Z}\to G$, it is enough to know $\varphi(1)$.
More precisely, if $\varphi$ is such a morphism, then, for all $m\geq 0$, we have $\varphi(m)=\varphi(1+\cdots+1)=\varphi(1)^m$.
If $m<0$, then $\varphi(m)=\varphi(-(-m))=\varphi(-m)^{-1}$ (a morphism repects inverses), so $\varphi(m)=(\varphi(1)^{-m})^{-1}=\varphi(1)^m$.
Finally, $\varphi=h_x$ with $x=\varphi(1)$.
In particular, if $G=\mathbb{Z}_n$, you get $n$ such morphisms.
Concerning your second question, you have only the trivial group/ring morphism. Any element of $\mathbb{Z}_n$ has finite order. But a morphism sends an element of finite order to an element of finite order. Since the only element of $\mathbb{Q}$ of finite order is $0$....
Edit Some answers were given while I was typing. Feel free to delete this post if it seems useless.
Best Answer
Your answer to the first question is correct.
For the second question, suppose that there were an onto homomorphism $f : \mathbb{Q} \to \mathbb{Z}$. Then there exists some $q \in \mathbb{Q}$ such that $f(q) = 1$. But then, $x = f(q/2)$ is an integer satisfying $$2x = x+x = f(q/2) + f(q/2) = f(q/2 + q/2) = f(q) = 1,$$ which is impossible. Therefore there is no such $f$.