Short Answers:
1.- Because $\,P_5\le N_G(P_5)\,$ and $\,|N_G(P_5)|=|P_5|\,$
2.- The "2nd Noether Theorem" seems to be what others (like me) call one (the second or the third, usually) of the isomorphism theorems, and the formula you want to use is precisely the order of the group
$$P_5P_3/P_3\cong P_5/(P_5\cap P_3)$$
otherwise you wouldn't be able to deduce $\,|P_3P_5|=15\,$ ...
3.- The group $\,\Bbb Z_{15}-\,$ the cyclic group of order $\,15\,$ is abelian , so it trivially normalizes its own subgroups...(this means $\,P_5\,$ is normal in $\,P_3P_5\,$ )
$P$ is certainly inside $Q\cap H$. On the other hand $Q\cap H$ is a $p$-group, being inside $Q$ (by Lagrange). Since $P$ is a maximal $p$-subgroup of $H$, you get $P=Q\cap H$.
Best Answer
In general if $H$ is a subgroup of $G$ and $g\in G$, then we have $|gHg^{-1}|=|H|$, so it follows that the conjugates of a Sylow subgroup are also Sylow subgroups.