On Conjugacy Classes of Alternating Group $A_n$

Hint for part 1: The product of two 3-cycles $\sigma=(123)(456)$ commutes with $\tau=(14)(25)(36)$. This is because $\tau (123)\tau^{-1}=(456)$ and $\tau(456)\tau^{-1}=(123)$. The permutation $\tau$ is odd, because it has an odd number of transpositions. Generalize to conclude that if a permutation has two cycles of an equal odd length then it commutes with an odd permutation.

Hint for part 2: Remember that the size of the conjugacy class of an element $x$ in a group $G$ is $|G|/|C_G(x)|$. Here we have both $S_n$ and $A_n$ assuming the role of $G$. But obviously $C_{A_n}(x)=C_{S_n}(x)\cap A_n$, and $C_{S_n}(x)$ consists either of only even permutations (when intersecting with $A_n$ won't change anything) or ...

Hint: sometimes it's hard to understand the general idea, and it might be a good idea to have an example in mind. Think of $A_4$. The cycle types of elements in $A_4$ are: $1111$(identity), $22,13$. The only cycle type that consists of distinct odd integers is $13$. That is, $K$ is a union of two conjugacy classes in $A_4$ iff $K$ has an element of the form $(- - -)$. Now, you can (easily) check that the elements $(123),(132),(124),(142),(234),(243),(134),(143)$ form two conjugacy classes in $A_4$.