If $HK$ is a cyclic normal subgroup of $G$, then any subgroup of $HK$ is characteristic in $HK$, and therefore normal in $G$ by this. It's not immediately clear to me how to prove this directly, but here's another approach.
First, there's a general fact that any group of order 2 mod 4 has a subgroup of index 2. Proof: embed $G$ in $S_{|G|}$ by the left-multiplication action, and intersect with $A_{|G|}$. This will have index 2 in $G$ unless $G$ is already contained in $A_{|G|}$. But it isn't: any order-2 element in $G$ acts as $|G|/2$ transpositions, which is an odd permutation.
So $G$ must contain a subgroup $L$ of order 1005, necessarily normal in $G$. If we can show that $L$ has a characteristic subgroup of order 5, then the link above proves that this subgroup is normal in $G$. To accomplish this, we can go back to Sylow's theorem, and use some of the ideas you already came up with for $G$.
First, Sylow's theorem tells us that the number of Sylow 5-subgroups of $L$ is either 1 or 201. If there is only one, then it's characteristic, so suppose for contradiction that there are 201. This would account for 804 nonidentity elements, which leaves only 201 elements of $L$ not of order 5. But by your argument, there is a unique Sylow 67-subgroup of $L$, and its product with any order-5 subgroup is a cyclic subgroup of order 335. This must have $\varphi(335) = 4 \cdot 66 = 264$ generators, which is too many. So $L$ must have a single Sylow 5-subgroup, which is therefore characteristic in $L$ and thus normal in $G$.
Here is a solution to part $a$. Note that $|G/N|=\frac{|G|}{|N|}=335$. Now assume $g\in G$ is an element of order $2$. Then $g^2=e\in N$ when $e$ is the identity element of $G$. So that way we get $g^2N=N$. And then:
$N=N^{168}=(g^2N)^{168}=g^{336}N=gN(gN)^{335}=gN$
Where we use the fact that $|G/N|=335$ and hence $(xN)^{335}=N$ for all $xN\in G/N$. So we got $gN=N$ and that means $g\in N$. Now use the same idea for elements of order $3$.
Part $b$: I don't really understand your explanation, and anyway the answer is wrong. I'll give a hint: $\mathbb{Z_6}$ and $S_3$ are all the groups of order $6$ up to isomorphism. Your group $N$ is abelian, so it is $\mathbb{Z_6}$. Now it should be easy to count the elements of order $2$ and $3$. There is one element of order $2$ and two elements of order $3$.
Now to part $c$. I'll leave to you as an exercise to prove the following lemma:if $P$ is a $p$-group, $X$ is a finite set and $P$ acts of $X$ then the number of elements of $X$ with orbit of size $1$ is congruent to $|X|$ mod $p$. Now we'll use it.
Let $P$ be a $3$-Sylow subgroup of $G$. As we already know there are only $2$ elements of order $3$ we also know this is the only $3$-Sylow subgroup and hence it is normal in $G$. Now let $Q$ be a $67$-Sylow subgroup of $G$ and define an action of $Q$ on $P$ by $g.x=gxg^{-1}$. It is an action because $P$ is normal in $G$. Now let $X$ be the set of elements of $P$ with orbits of size $1$. These are the elements of $P$ that commute will all the elements of $Q$. You can check it is a subgroup of $P$, so its order must be $1$ or $3$. But $Q$ is a $67$-group and hence by the lemma $|X|$ is congruent to $|P|=3$ mod $67$. So $|X|$ can't be $1$, so it must be $3$, which means $X=P$. Every element of $P$ commutes with every element of $Q$ and that means $Q\leq C_G(P)$. By Lagrange's theorem we get $67$ divides $|C_G(P)|$.
Now do exactly the same process with a $5$-Sylow subgroup and conclude that $5$ divides $|C_G(P)|$. Also, $P$ is abelian and hence $P\leq C_G(P)$, so also $3$ divides $|C_G(P)|$. Finally, as there is only one element $g\in G$ of order $2$ we can conclude that this element is in the center of $G$ and hence $\langle g \rangle\leq C_G(P)$ so also $2$ divides $|C_G(P)|$. So as $2,3,5,67$ all divide $|C_G(P)|$ we get that $2010$ divides it too, so $C_G(P)=G$. That means $P$ is in the center of $G$.
Alright, so now look at the group $Z(G)\cap N\leq N$. The identity is there, the only element of order $2$ is there and as we showed both elements of order $3$ are there. So this group contains more than half of the elements of $N$ and by Lagrange's theorem we get that $Z(G)\cap N=N$. And that implies $N\leq Z(G)$.
Best Answer
Here is an outline. Let $G$ be a group of order $2010$, and let $N$ be a normal subgroup of $G$ with $|N| = 6$.
You need to show that $G$ has a normal $5$-Sylow, so let $R \leq G$ be a Sylow $5$-subgroup of $G$. Prove that $RN$ is a subgroup of order $30$, and that $R$ is normal in $RN$.
Thus $N_G(R)$ has order divisible by $30$. Apply Sylow's theorem to conclude that $N_G(R) = G$ and $R$ is normal in $G$.