I have that $G$ is a simple group of order $360$. By the 2nd, 3rd Sylow theorems we know that there are $n_3 = 1+3k, k \in \mathbb{Z}$ Sylow 3-subgroups and that $n_3$ divides $2^3 \cdot 5$, since $360 = 2^3 3^2 5$.
This gives us that $k = 0, k = 1, k = 3, k = 13$. We can immediately rule out $k = 0$ since $G$ is simple. How do I rule out the possibilities $k = 1, k = 13$ and show that their pairwise intersection is the singleton set containing $1$?
edit: I believe $k = 1$ can be ruled out since $1$ is not a prime
Here is a sketch proof. You need to complete the details yourself.
Suppose two distinct Sylow 3-subgroups $P,Q$ have nontrivial intersection $H$, so $|H|=3$. Then $C_G(H)$ contains $P$ and $Q$, and has at least $4$ Sylow $3$-subgroups. If it had more than $4$ then its index in $G$ would be too small so it has exactly 4 and $|C_G(H)|=36$ with $C_G(H)/H \cong A_4$. So $C_G(H)$ has a normal Sylow $2$-subgroup $T$ of order $4$ and $N_G(T)$ contains both $C_G(H)$ and a Sylow $2$-subgroup of $G$, so $|N_G(T)| \ge 72$ and its index in $G$ is too small.
So the Sylow $3$-subgroups have trivial intersection. If there were 40 of them, then there are only $39$ elements of $G$ outside of their union.
If $n_5=6$, then $G \cong A_6$, which we know has exactly 10 Sylow $3$-subgroups. So $n_5 \ge 11$, and there are too many elements of order $5$. So we must have $n_3=10$.