I'll try

Edited Additions: You can also let $A_j$ be the statement "$\aleph_{j}^{L}$ is a cardinal (in $V$)" (i.e., the $j^{th}$ uncountable cardinal of the constructible universe is a cardinal in the actual universe). In this case, you could simultaneously force over $L$ (via a countable product of posets from $L$ collapsing cardinals) to think about add a surjection from $\aleph_{j-1}^L$ to $\aleph_{j}^L$ exactly when $s_j = 0$ so that the other questions latercardinal $\aleph_{j}^L$ becomes an ordinary ordinal of size $|\aleph_{j-1}^L|$ in the forcing extension. In the case that all of the $s_j$'s are $0$, the first $\aleph_0$ many cardinals of $L$ all become countable ordinals from the perspective of the forcing extension whereas if they're all $1$'s, then we have done trivial forcing and so the forcing extension is $L$.

Now after showing the desired relative consistency results as above, you can note (for your If so part) that you are only considering countable ordinals here from the perspective of most universes. For example, if a certain type of Real exists in your universe, mainly $0^{\sharp}$, then the true $\aleph_1$ will be inaccessible in $L$ and more so all of the $\aleph_j^L$'s for $j \in \mathbb{N}$ will be very puny countable ordinals in the said universe. Of course, this is probably cheating, but I thought I'd mention it anyway.

Also to your subquestion, $2^{\aleph_0} = \aleph_1$ and $2^{\aleph_0} = \aleph_2$ are very meaningful distinctions. But also under ZFC, $2^{\aleph_0}$ needs to be quite large in order to extend the Lebesgue measure to a countably additive measure on the full powerset of $\mathbb{R}$.

(1) Yes, let $A_j: 2^{\aleph_j} \neq \aleph_{j+1}$ (i.e., GCH does not hold at $\aleph_j$). We can do this by simultaneously forcing (via a countable product of posets adding Cohen subsets) $2^{\aleph_j} = \aleph_{j+1+s_j}$ where $s_j$ represents the truth value at $j$.