If $G/N$ and $H/N$ are isomorphic, does it imply $G$ and $H$ are isomorphic? If $N$ is a subgroup of both $G$ and $H$, and $G/N \cong H/N$, then is $G \cong H$?  (Or more generally, if $N$ and $M$ are subgroups of $G$ and $H$ respectively, and $N \cong M$ and $G/N \cong H/M$, then is $G \cong H$?)
I suspect the answer is no, but if the answer is yes, is there a natural isomorphism between $G$ and $H$ given an isomorphism $\phi:G/N \rightarrow H/N$?
Also, just making sure, is $(G \times H)/G \cong H$ and $(G \times H)/H \cong G$?
 A: Your first/second question can be reformulated as asking whether the isomorphism class of the normal subgroup $N$ of the group $G$, and that of the quotient group $G/N$, together determine the isomorphism class of $G$.
The short answer is no, as shown by @Jim. A fuller answer is that you are actually inquiring about the intricate problem of group extensions.
As to your third question, I recommend a bit of care. If you write
$$
(\mathbf{Z} \times \mathbf{Z}) / \mathbf{Z},
$$
the $\mathbf{Z}$ at the denominator refers to which of the many copies of $\mathbf{Z}$ in $\mathbf{Z} \times \mathbf{Z}$? If you choose for instance
$$
\{ (2x, 0) : x \in \mathbf{Z} \} \cong \mathbf{Z}
$$
as your copy of $\mathbf{Z}$ at the denominator, then the quotient is definitely not isomorphic to $\mathbf{Z}$, but to $\mathbf{Z}/2 \mathbf{Z} \times \mathbf{Z}$.
So you should probably say something like
$$
1 \to H \to G \times H \to G \to 1
$$
is an exact sequence, where $H \to G \times H$ is the map $h \mapsto (1, h)$ and $G \times H \to G$ is $(g, h) \mapsto g$.
A: $N = \mathbb Z/2$ is isomorphic to a subgroup of $G = \mathbb Z/2 \times \mathbb Z/2$ and of $H = \mathbb Z/4$.  These groups are not isomorphic but $G/N \simeq \mathbb Z/2 \simeq H/N$.
A: Let $G=S_3$, $H=\mathbb Z_6$, $N_1=\langle(1~2~3)\rangle$ and $N_2=\langle[2]\rangle$. We have $$N_1\leq G,~~~N_2\leq\ H,~~~N_1\cong N_2$$ and $G/N_1\cong H/N_2$ but $G\ncong H$.
