How do I know if I can make counting statements for sylow theorems that rely on intersecting trivially? I.e it is a common argument that I use in which I rely on we have $x$ amount of sylow $p$-groups that are cyclic, if they intersect trivially they give $y$ number of non-trivial elements and this forces $n_q=1$ which gives us a normal subgroup, simple-non simple yada yada, or this normal subgroup is cyclic, abelian, solvable.
When is it valid to argue using my 'trivial intersection -> so many distinct elements'?
Example for clarity: $|G|=36=3\times12=3^2\times 2^2$, so in the case that $n_3=4$, we have $4$ sylow $3$-groups of order $9$, if they intersect trivially we have $(9-1)\times 4$ non-trivial elements i.e. $32$ non trivial elements, we need to obtain $3$ more non-trivial elements, this forces $n_2=1$, which gives us a normal subgroup(since $n_2=1$) and this is cyclic of order $2$, and cyclic means abelian which means solvable.
In the example you give, the Sylow subgroups might not intersect trivially, so you also have to consider the situation when they do not.
In that case, two Sylow $3$-subgroups $P_1$ and $P_2$ would intersect in a subgroup $Q$ of order $3$. We can analyse that case by considering the centralizer $C_G(Q)$ of $Q$ in $G$. Since it contains both $P_1$ and $P_2$, $C_G(Q)$ has more than one Sylow $3$-subgroup, so it must have at least $4$, but then we get $|C_G(Q)| \ge 36 = |G|$, so $C_G(Q) = G$, and hence $Q \lhd G$ and we can carry on the analysis by studying $G/C_G(Q)$.