Is there a Cantor-Schroder-Bernstein statement about surjective maps?
Solution 1:
For the first one the need of the axiom of choice is essential. There are models of ZF such that $A,B$ are sets for which exists surjections from $A$ onto $B$ and vice versa, however there is no bijection between the sets.
Using the axiom of choice we can simply inverse the two surjections and have injections from $A$ into $B$ and vice versa, then we can use Cantor-Bernstein to ensure a bijection exists.
The second one, I suppose should be $f\colon A\to B$ injective and $g\colon A\to B$ surjective, again we need the axiom of choice to ensure that there is a bijection, indeed there are several models without it where such sets exist but there is no bijection between them. Using the axiom of choice we reverse the surjection and use Cantor-Bernstein again.
It should be noted that without the axiom of choice it is true that if $f\colon A\to B$ is injective then there is $g\colon B\to A$ surjective. Therefore if the first statement is true, so is the second, and if the second is false then so is the first.
Another interesting point on this topic is this: The Partition Principle says that if there is $f\colon A\to B$ surjective then there exists an injective $g\colon B\to A$. Note that we do not require that $f\circ g=\mathrm{id}_B$, but simply that such injection exists.
This principle implies both the statements, and is clearly implied by the axiom of choice. It is open for over a century now whether or not this principle is equivalent to the axiom of choice or not.
Lastly, as stated $f\colon A\to B$ injective and $g\colon B\to A$ surjective cannot guarantee a bijection between $A$ and $B$ with or without the axiom of choice. Indeed the identity map is injective from $\mathbb Z$ into $\mathbb R$, as well the floor function, $x\mapsto\lfloor x\rfloor$ is surjective from $\mathbb R$ to $\mathbb Z$ but there is no bijection between $\mathbb Z$ and $\mathbb R$.
Solution 2:
As stated, 2. is clearly false (just take $A=\{ 0,1\} ,B=\{ 0\}$ with $f$ identically zero, and $g$ likewise). I will assume that it's actually $f:A\to B$ and $g:A\to B$.
Using axiom of choice, both statements can be shown to be true, simply because when we have a surjection $f:A\to B$, then by axiom of choice we can choose a right inverse $f^{-1}:B\to A$ which will be injective, so we can reduce both statements to the usual C-B-S.
Without choice, neither statement can be proved.
For the first one, see https://mathoverflow.net/questions/38771 (apparently, it would imply countable choice).
For the second one, see https://mathoverflow.net/questions/65369/half-cantor-bernstein-without-choice.