Let $H$ be a 7 Sylow subgroup of $A_{20}.$ Then prove that $H$ is not normal in $A_{20}$ and any 7 Sylow subgroup of $S_{20}$ is contained in $A_{20}.$
Proof:
since $7^2$ divides order of $A_{20}$ but $7^3$ does not divide, we have $|H|=49.$ Now there are more than $2 \times6!=2 \times720$ elements of order $7$ in $A_{20}$, because $(1\ 2\ 3\ 4\ 5\ 6\ 7)$ and $(8\ 9\ 10\ 11\ 12\ 13\ 14)$ are of order $7$ in $S_{20}$ and they are even permutations so contained in $A_{20}$. If $H$ is a normal 7 Sylow subgroup then it must be unique in $A_{20}$. And every element of order $7$ must be contained in $H$ (By Sylow's second theorem), which is not possible because $H$ can at most contain $49$ elements. So $H$ is not unique. Am I correct in this part?
In second part, every element of order of 7 Sylow subgroup is of order $7$ or $1$. So it must be even permutation. So it is contained in $A_{20}$. Am I correct? Thank you.
Your approach seems fine to me, though your first argument is a bit cluttered. They key observation you have made is that if $H$ is normal in $A_{20}$, then the elements of $A_{20}$ of order $7^k$ ($k\in\{0,1,2\}$) form a subgroup of order $49$. So indeed it suffices to show that there are more than $49$ elements of order $7^k$ in $A_{20}$. But I do not understand the argument you give for the existence of $2\times6!=2\times720$ elements of order $7$ in $A_{20}$.
My first thought would be to note that any choice of $7$ elements from $S_{20}$ yields $6!$ distinct $7$-cycles in $S_{20}$, yielding at least $\binom{20}{7}\times6!>49$ elements of order $7$. (There are even more though.)
Your idea for the second part is good, though the implication $$\text{ the order of $\sigma$ is $7$ or $1$}\qquad\Rightarrow\qquad \sigma\in A_{20},$$ requires a (simple) argument.