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 ...
I don't think there is a very conceptual approach giving a meaning to the comparison of conjugacy class sizes, but I think there is a way to make the argument quite short. The first thing to do is to cancel the term $a_1!$ for $i=1$ (for the points remaining fixed) in the denominator against a part of the numerator $n!$, which leaves the numerator equal to $n(n-1)\ldots(a_1+1)$ (the number of factors is the sum of the sizes of the cycles for the conjugacy class, in the group-theoretic sense where cycles cannot have length$~1$).
A first easy simplification rids us the case of mixed cycle structures, classes with is more than one different cycle length${}\geq2$. For such cycle types we can map permutations in the class to another non-trivial cycle type by simply omitting the cycle(s) of the longest length$~l$ from the decomposition. This map commutes with any conjugation by a permutations (which simple relabels elements in the cycles) and is therefore surjective to a single new conjugacy class; moreover it is not injective since there is always more than one way to reconstruct the length$~l$ cycle(s). Therefore such classes are never the smallest non-trivial ones.
The next step is to show that we need only deal with the case of transpositions, possibly flying in formation. More precisely (and with three exceptions that do not occur for $n\geq7$), the class with $m$ disjoint $k$-cycles with $k>2$ is larger than the one with $m$ disjoint $2$-cycles. We have $a_k=m$ and factor our formula as
$$
\frac{n(n-1)\ldots(n-mk+1)}{k^m}\times\frac1{m!},
$$
where the second factor is identical to the one for $k=2$. In the first factor replacing $k$ by $2$ means dropping the last $m(k-2)$ factors from the numerator and dividing the denominator by $(k/2)^m$. Using that a product of $k-2$ distinct positive integers can only be${}\leq k/2$ if $k\leq4$, we see that this decreases the first factor except when $a_1=0$, and $(m,k)\in\{(1,3),(1,4),(2,3)\}$, which happens only for $n=3,4,6$ (it explains that $3$-cycles are least numerous for $n=3$, that for $n=4$ there are as many $4$-cycles as $2$-cycles, namely $6$, and that for $n=6$ there are fewer ($40$) pairs of $3$-cycles then pairs of transpositions $(45)$).
We have reduced to the case of $m$ disjoint transpositions, which we shall compare with isolated transpositions ($m=1$). For fixed $m$, increasing $n$ leaves the denominator unchanged and increases all factors in the numerator by$~1$. This gives more relative increase for $m>1$ than for $m=1$, so if the class with $m$ disjoint transpositions is to be less numerous than that with one transposition, it must already be the case for the minimal possible value of $n$, which is $n=2m$. The formula for the number of products of $m$ disjoint transpositions when $n=2m$ is the product $1\times 3\times\cdots\times(2m-1)$ (which some people, defying the ambiguity, write as $(2m-1)!!$) which is${}\leq\binom{2m}2$ only for $m=2,3$; this explains these cases for $n=4,6$. For $n\geq7$ this does not happen any more.
Best Answer
Let $\sigma$ be an even permutation. If there is an odd permutation commuting with $\sigma$, then $\sigma$ has the same conjugacy class in $A_n$ as in $S_n$. Otherwise the centraliser of $\sigma$ in $S_n$ is a subgroup of $A_n$ and the conjugacy class of $\sigma$ in $S_n$ splits into two classes in $A_n$.
If $\sigma$ has a cycle of even length, then that's an odd permutation centralising $\sigma$. If $\sigma$ has two cycles of the same odd length $k$, then there is an involution swapping these two cycles, and centralising $\sigma$. This involution has $k$ two-cycles and so is an odd permutation.
The only remaining possibility is if the cycle decomposition of $\sigma$ consists of cycles of distinct odd lengths. In this case the centraliser of $\sigma$ is generated by these cycles and so is contained in $A_n$. These are the conjugacy classes of $S_n$ that split into two classes in $A_n$.