I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way:
Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups.
Similarly, the Sylow-3 subgroups must be cyclic of order 3. There are 20 3-cycles in $S_5$, 2 to a subgroup, so 10 Sylow-3 subgroups.
Since 8, but not 16, divides 120, the Sylow-2 subgroups must have order 8. Now, $S_4$ contains three copies of the dihedral group of order 8, and $S_5$ contains 5 copies of $S_4$, so I get 15 Sylow-2 subgroups.
Something like this ought to work for $A_5$.
EDIT: I think OP wants me to elaborate on the dihedral-group part of the argument.
Take a square, label its vertices, cyclically, with 1, 2, 3, 4. Then the element $(1234)$ of $S_4$ has a natural interpretation as the rotation, one-fourth of the way around, of the square, and the element $(13)$ is the flip in the diagonal through 2 and 4, so these two elements of $S_4$ generate a subgroup isomorphic to the dihedral group of order 8.
The same is true for the elements $(1342)$ and $(14)$, and also for the elements $(1423)$ and $(12)$, and those are the three copies of the dihedral group in $S_4$.
Now if you pick any one of the numbers 1, 2, 3, 4, 5, and consider all the elements of $S_5$ that fix that number, you get a subgroup of $S_5$ isomorphic to $S_4$. Those are your five copies of $S_4$ in $S_5$.
You could argue as follows, since Sylow subgroups are permuted transitively under conjugation.
If the number of Sylow $3$-subgroups is $4$, then they are permuted transitively by conjugation, and this gives rise to a non-trivial homomorphism to $S_4$. But $S_4$ has order $24\lt 60$ so such a homomorphism would have a non-trivial kernel.
Best Answer
Sylow's third theorem says that if $|G|=p^na$ where $p$ and $a$ are coprime and $p$ is prime, then the number of Sylow $p$-subgroups (those are subgroups of order $p^n$) denoted by $n_p$ satisfy $n_p\equiv1(\textrm{mod} \hspace{3pt}p)$ and $n_p$ divides $a$. So in your case, $n_3$ divides $20$ and $n_3\equiv 1(\textrm{mod} \hspace{3pt}3)$
We have $20$ elements of order $3$ and not $20$ subgroups of order $3$. Since each Sylow 3-subgroup will have two non-identity elements we have $\displaystyle\frac{20}{2}=10$ Sylow-3 subgroups in $A_5$.