Cardinal of $V_{\omega+\alpha}$ I do not understand why $card(V_{\omega+\alpha})=\beth_\alpha$. The steps in the recursion for $0$ and succesor ordinals are quite easy, but I do not manage to prove it for limit ordinals.
I see that if $l$ is a limit ordinal, we have $\beth_l=\cup_{i<l}\,\beth_i$ and $V_{\omega+l}=\cup_{i<l}\,V_i$. By the recursion hypothesis, for all $i<l$, we a have a bijection between $V_{\omega+i}$ and $\beth_i$. But I do not see why we should have one between $V_{\omega+l}$ and $\beth_l$ (see this to understand my difficulties).

The main difference between this situation and the situation in your previous question is that here we will additionally have that $|A| < |A^\prime|$ (and $|B| < |B^\prime|$), and, furthermore, $A^\prime$ (and $B^\prime$) are infinite sets.  In this situation we can be guaranteed that $| A^\prime \setminus A | = | B^\prime \setminus B |$ (again, provided $|A| = |B|$ and $|A^\prime| = |B^\prime|$).  In fact, we can say even more:

If $A \subseteq A^\prime$ are such that $A^\prime$ is infintie, and $|A| < |A^\prime|$, then $|A^\prime \setminus A | = | A^\prime |$.

This stems from the following fact of cardinal arithmetic:

If $\kappa \leq \lambda$ are cardinals such that $\lambda$ is infinite, then $\kappa + \lambda = \lambda$.


Going slowly, if $\alpha$ is a limit and we know that $| V_{\omega+\xi}| = \beth_\xi$ for each $\xi < \alpha$ we can first define for each $\xi < \alpha$ the set $$W_{\xi} = V_{\omega+\xi+1} \setminus V_{\omega_\xi}.$$  It is easy to see that $V_{\omega+\alpha} = \bigcup_{\xi < \alpha} W_\xi$, and that these sets are pairwise disjoint.  Furthermore by the induction hypothesis (and the above) $| W_{\xi} | = \beth_{\xi+1}$ for each $\xi < \alpha$.  Now calculate $| V_{\omega+\alpha} |$ via $$| {\textstyle \bigcup_{\xi < \alpha}} W_\xi | =  {\textstyle \sum_{\xi < \alpha}} | W_\xi |.$$
