Yes, this is possible. If $M_1^\#$ exists and is $\omega_1$-iterable then $x^\#$ exists for all reals $x$: By genericity iterations (either Woodin's or Neeman's version), for any real $x$ there is a countable iterate $N$ of $M_1^\#$ so that $x$ is generic over $N$ for a forcing of size the Woodin cardinal of $N$. So $N[x]$ still "has a sharp" and $x^\#$ exists. Neeman's version accomplishes this by an iteration of length $\omega$, so the same conclusion holds in all generic extensions of $V$ in which $(M_1^\#)^V$ is still $\omega+1$-iterable (i.e. admits cofinal wellfounded branches for iteration trees of length $\omega$).

If now, e.g. $M_1^\#$ exists, is $\omega_1$-iterable and $V=L[A]$ for a set $A$, then in the extension by $\mathrm{Col}(\omega, A)$, $(M_1^\#)^V$ cannot be $\omega+1$-iterable.

Yes, this is possible. If $M_1^\#$ exists and is $\omega_1$-iterable then $x^\#$ exists for all reals $x$: By genericity iterations (either Woodin's or Neeman's version), for any real $x$ there is a countable iterate $N$ of $M_1^\#$ so that $x$ is generic over $N$ for a forcing of size the Woodin cardinal of $N$. So $N[x]$ still "has a sharp" and $x^\#$ exists.

If now, e.g. $M_1^\#$ exists, is $\omega_1$-iterable and $V=L[A]$ for a set $A$, then in the extension by $\mathrm{Col}(\omega, A)$, $(M_1^\#)^V$ cannot be $\omega_1$-iterable.