Describe all homomorphisms of $S_n$ into $S_3$, $n \geq 5$.
Somehow it is a bit unclear to me what describing means. However, I would say that describing is meant with respect to generators. We have $$S_n = \left\langle \begin{pmatrix}1 & 2\end{pmatrix},\begin{pmatrix}1 & 2 & \dots & n\end{pmatrix}\right\rangle$$ and $|S_3| = 6$. So theoretically there would be $36$ possible homomorphisms. But thats not really descriptive. In the former part of the exercise we've already proved that $\text{Aut}(S_3) \cong S_3$ but I do not really know if this helps.
The trick here is to recall that the kernel of a group homomorphism is a normal subgroup. If $\phi:S_n \rightarrow S_3$ is a group homomorphism, we must have $S_n / \ker(\phi) \cong \operatorname{Im}(\phi)$, were $\ker(\phi)$ is normal in $S_n$.
For $n \geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$, and $S_n / A_n \cong \mathbb{Z}_2$ (which is a subgroup of $S_3$, namely, a subgroup generated by a transposition, so a homomorphism $S_n \rightarrow S_3$ is possible).
Lastly, don't forget about the trivial homomorphisms ;)
P.S. if you want to consider homomorphisms $S_4 \rightarrow S_3$, it's useful to know that $S_4$ also has a normal subgroup isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. Given that $S_4 / \mathbb{Z}_2 \times \mathbb{Z}_2$ has the same order as $S_3$, there is a possibility of an additional (surjective) homomorphism. Note that it suffices to check whether this quotient group is abelian.