= Solution
The <Cantor-Schröder-Bernstein theorem> states that <injections> $f:A\to B$ and $g:B\to A$ imply the existence of a <bijection> $A\to B$. Here is the injection-chain <formal proof>, so no fixed-point theorem is needed. Define
$$
A_0=A\setminus g(B),\qquad A_{n+1}=g(f(A_n)),\qquad C=\bigcup_{n<\omega}A_n,
$$
and <set>
$$
h(a)=\begin{cases}f(a),&a\in C,\\ g^{-1}(a),&a\notin C.\end{cases}
$$
The second branch is defined because $a\notin C$ implies $a\notin A_0$, hence $a\in g(B)$, and the inverse there is unique by <injectivity> of $g$. Each branch is <injective>. A collision between branches would have $f(c)=g^{-1}(a)$ with $c\in C$ and $a\notin C$, implying $a=g(f(c))\in C$, a contradiction. Thus $h$ is <injective>.
For <surjectivity> let $b\in B$. If $g(b)\notin C$, then $h(g(b))=b$. If $g(b)\in C$, it is not in $A_0$, so it belongs to some $A_{n+1}=g(f(A_n))$. There is $a\in A_n$ with $g(b)=g(f(a))$; <injectivity> of $g$ gives $b=f(a)=h(a)$. This proves that $h$ is a <bijection> and completes the theorem without the <axiom of choice>.
For the comparison of <cardinalities> take the <power set> of $\omega$ as the reference <set>. Its <cardinality> is by definition
$$
\boxed{|\mathcal P(\omega)|=2^{\aleph_0}.}
$$
The remaining parts give explicit <injections> or <bijections> comparing each requested <set> with this one.
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