Cocycles and group extensions

Question

I'm trying to understand how elements in the second cohomology group with coefficients in some other group correspond to group extensions.

This is what I understand:

Suppose we have two (countable) abelian groups $G,H$. The second cohomology group $H^2(G,H)$ is defined as the group of all functions (cocycles) $c:G\times G\rightarrow H$ with the property that for all $g_1,g_2,g_3\in G$ $$c(g_1+g_2,g_3)+c(g_1,g_2) = c(g_2,g_1+g_3)+c(g_1,g_3)$$ Modulo functions (coboundaries) of the form $c(g,h) = f(g+h)-f(g)-f(h)$ for any $f:G\rightarrow H$.

Now any element $c\in H^2(G,H)$ defines a group extension $G\times_c H$ which is defined as the cartesian product of $G$ and $H$ with the multiplication $$(g,h)+ (g',h') = (g+g',c(g,g')+h+h')$$

If $c$ is a coboundary then one can prove easily that $G\times_c H$ is isomorphic to $G\times H$ (the usual product) with the isomorphism $(g,h)\mapsto (g,f(g)+h)$.

Note moreover, that the extension is abelian if $c(g_1,g_2) = c(g_2,g_1)$ (in fact, if there exists a co-boundary $d$ such that $c+d$ satisfies the above).

I have the following questions

1. I showed that if $c$ is a coboundary then $G\times_c H\cong G\times H$. More generally one can show that if $c-c'$ is a coboudnary then $G\times_c H \cong G\times_{c'} H$. Is this an if and only if? I mean, if $G\times_c H \cong G\times_{c'} H$ does it mean that $c-c'$ a coboundary? [As far as I understand this should be true, but I don't manage to prove it]

2. It is well known that the group $\mathbb{Q}$ is injective (because it is divisible). This means that any abelian extension of $\mathbb{Q}$ splits. In other words, if the answer for $1$ is yes, then this would imply that any abelian cocycle $c:G\times G\rightarrow \mathbb{Q}$ is a coboundary. Is there any explicit proof for that?

I'm more interested in question number 2, which technically can be formulated without the notion of a cocycle. Is any function $c:G\times G\rightarrow \mathbb{Q}$ with $c(g_1,g_2)=c(g_2,g_1)$ and $$c(g_1+g_2,g_3)+c(g_1,g_2) = c(g_2,g_1+g_3)+c(g_1,g_3)$$ takes the form $c(g_1,g_2)=f(g_1+g_2)-f(g_1)-f(g_2)$?