If $K$ is a normal subgroup of $G$ of order $4$, you may well argue in $H = G/K$, and reduce to show, via the correspondence theorem, that in a group of order $2 \cdot 3 \cdot 7 = 42$ there must be a normal subgroup of order $7$.
To do this, just consider that the number of $7$-Sylow subgroups in $H$ must divide $42/7 = 6$, and be congruent to $1$ modulo $7$.
$|G|=3\cdot5\cdot7$
$$ n_5 \in\{1, 21\} \; \text{&} \; n_7 \in \{1,15\}$$
$$ \text{If} \; n_5>1 \; \text{and} \; n_7>1 $$
$$\Rightarrow n_5=21 \; \text{&} \; n_7=15 $$
$$\Rightarrow \text{There are } \; 21(5-1) + 15(7-1) >105\; \text{elements of order 5 and 7 together, which is absurd.} $$
$$ \text{Hence either } \; n_5=1 \; \text{or} \; n_7=1$$
$$\text{If P and Q are Sylow 5 and 7 subgroups respectively then, one of the two has to be normal in G}$$
$$\text{Without loss of generality, assume P is normal in G }$$
$$P\lhd G \;\text{and} \;Q \leq G \Rightarrow PQ \leq G $$
$$\;\text{Futhermore}, P\cap Q=\{e\} \Rightarrow |PQ|=\frac{|P||Q|}{|P\cap Q|}=35\; $$
$$[G:PQ]=3, \;\text{which also happens to be the least prime dividing order of G, hence}\; PQ\lhd G$$
$$|PQ|=5.7 \; \text{ and 5 doesn't divide (7-1), so } \; PQ\; \text{is cyclic} $$
$$\Rightarrow \;\text{P and Q are characteristic in PQ. Since PQ is normal in G, we have both P and Q normal in G}$$
Best Answer
Answer with hints that must be checked and solved:
As already proved, a group $\;G\;$ of order $\;30\;$ must have either one unique subgroup of order five or one unique subgroup of order three.
Take now the unique such subgroup, say $\;N\;$ , and one of the (possibly several) groups of the other order, say $\;H\;$ . Since clearly $\;N\cap H=1\;$ , we get that $\;NH\;$ is a subgroup (why?) of order $\;15\;$ and thus $\;NH\lhd G\;$ (why?).
But a group of order $\;15\;$ is necessarily cyclic (why?), and since any subgroup of a normal cyclic subgroup of a big group is normal in the big group (why?), we get that both $\;N,H\lhd G\;$