For a subgroups $H_1$ and $H_2$ of $G$ and a normal subgroup $N$ of $G$: $H_1/N \cong H_2/N \Rightarrow H_1 \cong H_2$. Is it true? I need to understand what can we deduce from knowing that:

Let $G$ be a group, $H_1, H_2$ its subgroups and $N$ it's normal subgroup contained in both $H_1$ and $H_2$. Assume
$H_1/N \cong H_2/N$

Can we really say that then $H_1 \cong H_2$? Or anything else related?
 A: The groups are not necessarily isomorphic. 
A good strategy to approach such a question is to consider the simplest case first.  
Thus, to make things simple let us consider $N$ is a cyclic group of prime order $p$ sitting inside a to be specified group. 
For any subgroup $H$ of order $p^2$ that contains $N$, we'll have $H/N$ is of order $p$ and thus also cyclic. 
Thus if what you ask about were true then all subgroups of order $p^2$ that contain $N$ would need to be isomporphic. 
Is there a good reason for this? Not really. There are two types of groups of order $p^2$, the cyclic one of order $p^2$ and the direct product of two cyclic one of order $p$. 
Can we find a group that contains both these types of subgroups and have non-empty intersection? 
Let us take the direct product of a cyclic group of order $p^2$ with a cyclic group of order $p$. Then we certainly have cyclic subgroups of order $p^2$ and also a direct product of cyclic groups of order $p$. 
For example, if we write the group elements as pairs and use additive notation, first coordinate order $p^2$ second $p$, then the group generated by $(1,0)$ that is $H_1=\{(0,0),(1,0),(2,0), \dots, (p^2 -1,0)\}$ is cyclic of order $p^2$, the one generated by $(p,0)$ and $(0,1)$ is a product of two cyclic groups of order $p$, call it $H_2$; and the intersection is  $N=\{(0,0), (p,0), (2p,0), (3p,0), \dots, (p(p-1), 0)\}$ cyclic of order $p$.
Then $H_1/N$ and $H_2/N$ are isomorphic (both cyclic of order $p$) while $H_1$ and $H_2$ are not. 
Various other examples of this form could be given. (The example discussed in comments is the special case $p=2$.)
