Questions about a group with an index 4 normal subgroup

Hint: 4th (Lattice) Isomorphism theorem.

$G/N$ is (isomorphic to) either $C_4$ or $C_2 \times C_2$. What are the subgroups of $C_2 \times C_2$ ?

Here is a proof of the results (1) and (2) based on tips given by lhf and jspecter.

We have $\vert G : N \vert = 4$, hence $G/N \cong C_4$ or $C_2 \times C_2$. If $G/N$ is not cyclic, then $G/N \cong C_2 \times C_2$ and (2) is clear from the multiplication table.

Let $\mathcal{G} = \{ A \mid N \leq A \leq G\}$ and les $\mathcal{N}$ be the set of all subgroups of $G/N$. Then the 4th isomorphism theorem states that $$ \phi : \mathcal{G} \to \mathcal{N}: A \mapsto A/N$$ is bijective. Thus $\vert \mathcal{G} \vert = \vert \mathcal{N} \vert = 4$. Whether $G/N \cong C_4$ or $G/N \cong C_2 \times C_2$, it is clear that $G/N$ contains a subgroup $H$ of index 2. Hence, there exists $K \leq G$ such that $\phi(K) = K/N = H$. By virtue of the 4th isomorphism theorem, we have $\vert G/N:K/N\vert = \vert G:K \vert = 2$ and (1) is proven.