To question 1, there is such a pair. This is a minor reworking of Ashutosh's example you linked.
Start with $L,$ add an $\omega_1$-sequence of random reals $X=\langle r_{\alpha}: \alpha<\omega_1 \rangle,$ then a Cohen real $c,$ then another $\omega_1$-sequence of random reals $Y=\langle s_{\alpha}: \alpha<\omega_1 \rangle,$ then another random real $t.$ Let $N=L[X][c][Y][t]$ be the final model. Identify the reals in $X$ and $Y$ as elements of $[0,1],$ and $t$ as a subset of $\omega.$ In $N,$$\mu^*(X)=0$ and $\mu^*(Y)=1.$
We now work in $L[X][Y][t],$ noting that $X,$$Y,$ and $t$ are mutually random over $L.$ Let $\langle b_n \rangle$ be the increasing enumeration of $t,$ and $\langle c_n \rangle$ the increasing enumeration of $\omega \setminus t.$
We use $t$ to interweave $X$ and $Y.$ Precisely, define $\langle u_{\alpha}: \alpha<\omega_1 \rangle$ by letting $r_{\omega \alpha + n} = u_{\omega \alpha + b_n}$ and $s_{\omega \alpha + n} = u_{\omega \alpha + c_n}.$ Applying permutation invariance of random forcing in $L[t],$ we have that $u$ is a random $\omega_1$-sequence of random reals over $L[t]$ and $t \not \in L[u].$
Let $M=L[u] \subset L[X][Y][t] \subset N,$ and let $A=\{2^{-n-1}(1+u_{\omega \alpha +n}): n<\omega, \alpha<\omega_1\}.$ Then $\mu^*(A)^M=1$ and $\mu^*(A)^N = \sum_{n \in \omega \setminus t} 2^{-n-1} \in (0,1).$
Suppose there is a partition $A = B \sqcup C$ in $M$ as in the description of question 1. Then $t = \{n<\omega: \mu^*( C \cap [2^{-n-1}, 2^{-n}])^M = 0\} \in M,$ contradiction.
I think "morally" this counterexample doesn't really violate your intuition, in that we still have in $N$ a partition of $A$ into one set of "preserved points" and another set of "nullified points." A natural follow-up question (which I don't know the answer to) is whether there can be $A \in M \subset N$ such that (i) $\mu^*(A)^M > \mu^*(A)^N$ and (ii) in $N,$ every point in $A$ is a Lebesgue density point of a minimal measurable cover of $A.$