A group of order $30$ has a normal $5$-Sylow subgroup.

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 $\,P_3P_5\,$ )

$o(G) = 30 = 2*3*5$

Since the number of $5$ - Sylow subgroups is of form, $N_G(5) = 5k +1 $ and $N_G(5) \Big | o(G)$,

$\implies N_G(5) = 1$ or $N_G(5) = 6$

Similarly, $N_G(3) = 1$ or $N_G(3) = 10$.

First we will show that, $5$ - Sylow subgroup or $3$ - Sylow subgroup is normal is $G$.

If $N_G(5) = 6$ and $N_G(3) = 10$, then,

There are $(6*4 = 24)$ elements of order $5$ and $(10*2 = 20)$ elements of order $3$.

So a total of $24+20 =44$ elements. Which is a contradiction.

Thus, $N_G(5) = 1$ or $N_G(3) = 1$ or both.

If $N_G(5) = 1 $, then we are done. If not,

We have $N_G(3) = 1$ and $N_G(5) = 6$, let $H_3$ and $H_5$ be $3$-sylow and $5$-sylow subgroups respectively. Since $N_G(3) = 1$, $H_3$ is normal.

It is easy to see that $H_3H_5$ is a subgroup in $G$ and $o(H_3 H_5) = 15$.

By the argument used earlier, $H_5$ is normal in $H_3 H_5$.

$\implies$ There are only $4$ elements of order $5$ in $H_3H_5$.

Thus, the remaining $20$ elements of order $5$ lie in $G - H_3H_5$, but $\Big | G- H_3 H_5 \Big | = 15$, a contradiction.

Hence $N_G(5) = 1$

QED