Homomorphisms from $\mathbb{Z} \rightarrow S_3$.

a) is fine. But for b) Note that $n\in\ker \phi$ if $g^n=1$ for $g:=\phi(1)$. As such $g$ can only have orders $2$ or $3$ (or of course $1$) in $S_3$, only $2\mathbb Z$ and $3\mathbb Z$ (and of course $\mathbb Z$ that was excluded) are possible kernels.

c) is fine - though in the light of b) you were probably supposed to take $n\mapsto (1\,2\,3)^n$ as one of the examples.

The nice thing about the group $\mathbb{Z}$ is that it is the free group generated by one element. That is, you can map a generator (1 or -1) of $\mathbb{Z}$ to any element and that defines a (unique) group homomorphism. One way of seeing the existence of such a homomorphism is as follows:

Let $G$ be a group and $g \in G$. Then the subgroup $\langle g \rangle$ of $G$ is cyclic and hence is isomorphic to a quotient of $\mathbb{Z}$. This gives a surjection $\mathbb{Z} \twoheadrightarrow \langle g \rangle$ which sends $1$ to $g$. Composing this with the inclusion $\langle g \rangle \hookrightarrow G$ yields a group homomorphism from $\mathbb{Z}$ to $G$ that sends $1$ to $g$.

The homomorphism $\mathbb{Z} \rightarrow G$ with $1 \mapsto g$ has zero kernel if and only if $g$ has infinite order. Otherwise the kernel is $n \mathbb{Z}$ where $n$ is the order of $g$. So the nonzero kernels for the group homomorphisms from $\mathbb{Z} \rightarrow G$ are in one-to-one correspondence with the orders of the non-identity elements in $G$.