Here’s an answer to the first question.
Counting the number permutations of each cycle type is just a matter basic combinatorics. Consider a $4$-cycle, for instance. You can start it with anything, so let’s start it with $1$: $(1xyz)$. You now have $3$ choices for $x$, $2,3$, or $4$; no matter which you choose, you’ve $2$ choices for $y$; and then $z$ is already determined, so the total number of possibilities is $3\cdot2=6$. The same reasoning shows more generally that there are $(n-1)!$ ways to form an $n$-cycle from a set of $n$ objects.
For the type $(3,1)$ there are $4$ ways to pick which element is fixed, and there are $(3-1)!=2$ ways to to form the other $3$ elements into a $3$-cycle, so there are $4\cdot2=8$ permutations of this type.
For the type $(2,2)$ there are $3$ ways to decide which of $2,3$, and $4$ gets paired with $1$ in a transposition, and that completely determines the permutation: the other two elements must make up the other transposition.
For the type $(2,1,1)$ there are $\binom42=6$ ways to choose the two elements forming the transposition, and that completely determines the permutation.
And of course the identity permutation is the only one of type $(1,1,1,1)$, since such a permutation must have four fixed points.
Added: Once you know that the members of each cycle type form a conjugacy class, it’s easy to see why a normal subgroup must be a union of conjugacy classes. Suppose that $N$ is a normal subgroup, and let $\pi\in N$. For each $\sigma\in S_4$ we have $\pi^\sigma\in N^\sigma=N$, so the entire conjugacy class of $\pi$ must be a subset of $N$.