Let $n\in\mathbb{N}, n\ge3$. For each $k\in\{1,2,...,n\}$ let $H_k=\{\sigma^k|\sigma\in S_n\}.$
At 1. I thought about using Lagrange's Theorem so $ord(H_3)|ord(S_n)=n!$ but that doesn't help in any way and it seems I'm out of ideas, same problem with 2. I could use some hints. Thank you!
On
Hint: Show that for all $r$, any power of an $r$-cycle consists of $a$ disjoint $b$-cycles (possibly all fixed points) for some $ab=r$. Use this to show that if $\sigma$ is a $k$-cycle for odd $k>1$, then there does not exist $\tau\in S_n$ such that $\tau^k=\sigma$.
On
For $H_3$ to be a subgroup of $S_n$ you need that $\operatorname{id}\in H_3$, and that for all $\sigma,\tau\in H_3$ also $\sigma\tau\in H_3$. The former is clearly true as $\operatorname{id}=\operatorname{id}^3$. So apparently there exists a pair $\sigma,\tau\in H_3$ such that $\sigma\tau\notin H_3$. Find a pair.
For the second part; for which odd numbers does the same argument hold?
For 1, suppose $H_{3}\leq S_{n}$. Then $(ij)=(ij)^{3}\in H_{3}$ for all $i<j$. So $H_{3}=S_{n}$. Then $\sigma^{3}=(123)$ for some $\sigma\in S_{n}$. If $n=3$, such $\sigma$ does not exist. So let $n\geq 4$. Let $a_{1}=\sigma(1),a_{2}=\sigma(a_{1})$. We have $a_{1}\neq 1$. Suppose $a_{1}=2$. Since $\sigma^{3}(1)=2$, we have $\sigma(a_{2})=2$, so $a_{2}=1$. Then $(12)$ is a cycle in the disjoint cycle decomposition of $\sigma$. So $\sigma^{3}(2)=1$, contradiction. Hence $a_{1}\neq 2$. Since $\sigma(a_{2})=2$, we have $a_{2}\neq 2$ and $a_{2}\neq 1$. Then $(1,a_{1},a_{2},2,\cdots)$ is a cycle in the disjoint cycle decomposition of $\sigma$. At least one of $a_{1},a_{2}$ is not $3$. But $a_{1},a_{2}$ are not fixed by $\sigma^{3}$, contradiction.