I'm having troubles with this.
Let G be a group of order $1365$. Is $G$ simple? Normally, we aim to find a single Sylow p-subgroup and since its normal, we get the results.
However, factoring $1365=3 \cdot 5 \cdot 7 \cdot 13$ gives me(for $n_p$ being the number of Sylow p subgroups):
$n_3=\{1,7,13,91\}$
$n_5=\{1,21,91\}$
$n_7=\{1,15\}$
$n_{13}=\{1,105\}$
I might have missed some, but nonetheless we can't be sure that any of these Sylow p subgroups are normal. Then perhaps the group is simple?
How can i solve this task?
As @lulu suggests, this is a situation where we should apply a simple counting argument. This is a common theme in Sylow-theorem exercises.
Suppose for contradiction that there is a simple group $G$ of order $1365$. For this group, we must have $n_3, n_5, n_7, n_{13} > 1$, so by your computations we get
$$n_3 \geq 7$$ $$n_5 \geq 21$$ $$n_7 = 15$$ $$n_{13} = 105$$
This means the group has $1260=105\times 12$ elements of order $13$, $90$ elements of order $7$, at least $84$ elements of order $5$, and at least $14$ elements of order $3$ (make sure you understand how to prove all of this!!)
In total, we have accounted for at least $1260 + 90 + 84 + 14 = 1448$ distinct elements of $G$. This is impossible, because $\lvert G \rvert < 1448$.