Show that if $|G| = 30$, then $G$ has normal 3-Sylow and 5-Sylow subgroups. 
Show that if $|G| = 30$, then $G$ has normal $3$-Sylow and $5$-Sylow subgroups.

Let $n_3$ denote the number of 3-Sylow subgroups and $n_5$ the number of $5$-Sylow subgroups. Then, by the third Sylow theorem, $n_3$ divides $10$ and $n_3 \equiv 1 \mod 3$. From these two, we see that $n_3 = 1$. This implies that $G$ has a normal 3-Sylow subgroup.
Similarly, let $n_5$ denote the number of $5$-Sylow subgroups. Then, by the third Sylow theorem, $n_5$ divides $6$, and $n_5 \equiv 1 \mod 5$. So, we can infer that either $n_5 = 1$ or $n_5 = 6$. I don't know how to proceed from here. Can someone please help me? (This is not homework, only self study)

This is more delicate and difficult than most problems about Sylow subgroups, and it must be tackled by relating the 5-sylow subgroups and the 3-sylow subgroups. In general such a problem in group theory at the undergraduate level is either going to be trivially easy, or require this technique.
As you established $n_3 = 1$ is one possibility but it was noted in the comments that $n_3 = 10$ is also possible. Suppose $n_5 = 6$ so that the $5$-sylows are not normal. Then they must be their own normalizers because the number of them is equal to the index of the normalizer of any one, and they are the unique subgroups of order $5$. However the fact that only $5$ divides $30$ and not $25$ allows us to know that all of the elements of order $5$ in this group are contained in a $5$-sylow, and that these sylows have trivial intersection. But that means there are $(5-1)*6 = 24$ elements of order $5$ in this group. But this means that there cannot possibly be $10$ subgroups of order $3$ otherwise there would be $20$ elements of order $3$ and that would give $47$ elements in the group which would be absurd.
So we know that the $3$-sylows would have to be normal in this situation. Then we have that if $H_3$ is a $3$-sylow and $H_5$ is a $5$-sylow then $H_5H_3$ is a subgroup because $H_3$ is normal. But then $H_5H_3/H_3 \cong H_5/(H_5 \cap H_3)$ and counting orders we know that $H_5H_3$ has order $15$. But there is a unique group of order $15$ (there are many ways to show this, but in our particular situation the fact that $H_5H_3 \cong \mathbb{Z}_{15}$ comes down to the fact that $H_5$ is normal by a sylow argument on this subgroup) and thus this group is abelian and this contradicts that $H_5$ is its own normalizer. Thus $H_5$ is normal.
Finally we can now count again. Since $H_5$ is normal we can take any one of these $H_3$ and do the same argument as the above to realize that $H_3$ has a normalizer of order at least $15$ and thus $H_3$ must be normal because it is supposed to have normalizer or index either $1$ or $10$.
Hope this helps.
