Why is some power of a permutation matrix always the identity?

There are only finitely many ways to permute finitely many things. So in the sequence $$P^1,\ P^2,\ P^3,\ldots$$ of powers of a permutation $P$, there must eventually be two powers that give the same permutation, meaning that $P^i=P^j$ for some $i>j\geq0$. Permutations are reversible so $P$ is invertible, hence $$P^{i-j}=P^iP^{-j}=P^j(P^j)^{-1}=I.$$

And yes, a $2\times2$-block means a $2\times2$-matrix here. The hint suggest to choose a $5\times5$-matrix that has a $2\times2$-matrix and a $3\times3$-matrix on its diagonal, and zeroes elsewhere.

Ok, here you go: Note that in a finite group every element has finite order, see for example here for a proof. This means in a finite group $G$ you can find a $n\in \mathbb{N}$ for every $g\in G$ s.t. $g^n=e$.

Now you have a group homomorphism $\varphi:S_n\to Gl_n$ via the following map: take the standardbasis $e_1,\ldots,e_n$ and an element $\sigma \in S_n$, then $\varphi(\sigma)=(e_{\sigma(1)},\ldots,e_{\sigma(n)})$, here I mean the matrix spanned by this vectors. You should check that this is indeed a group morphism.

For a group morphism you have that $\varphi(\sigma)^n=\varphi(\sigma^n)$ and since $S_n$ is a finite group you find a $n \in \mathbb{N}$ s.t. $\varphi(\sigma)^n=\varphi(\sigma^n)=\varphi(id_{S_n})=id_{Gl_n}$.

Now for your last part I suggest you try the matrix that is associated to $(123)(45)$ under the above group morphism.