The answer to Question 2 is also "yes". Trivially so if $B\subseteq B'$ or $B'\subseteq B$, so assume otherwise. Given the affirmative answer to Question 1, it suffices to show that $mx=x$ when $mb=b$ for all $b\in B\cap B'$ (on the assumption that $B$ and $B'$ both fix $x$). Case 1: Assume that for some $a\in A$, $a\in B$ if and only if $a\in B'$. For any $c\in A, C\subseteq A$, let $g(C,c)\in M$ be the function that maps every member of $C$ to $c$ and every other member of $A$ to itself. Let $n=g(B'\backslash B,a)$, $n' =g(B\backslash B',a)$, $n''= g(A\backslash(B\cup B'),a)$, and $n'''(\{a\},m(a))$. $n$ maps every member of $B$ to itself, $n'$ maps every member of $B'$ to itself, and $n''$ and $n'''$ do both. So $nx = n'x=n''x=n'''x = x$. Since $m\circ n''\circ n'\circ n = n'''\circ n''\circ n'\circ n$, it follows that $mx = x$. Case 2: Suppose $B = A\backslash B'$ and $|A|> 2$. Consider some $b\in B,b'\in B'$. Since $B$ fixes $x$, so does $B\cup\{b'\}$; by Case 1, so does $(B\cup\{b'\})\cap B' = \{b'\}$; similarly, so does $\{b\}$; since $|A|>2$, there is an $a\in A$ such that $a\not\in \{b\}$ and $a\not\in \{b'\}$; so, by Case 1, $\{b\}\cap\{b'\}=\emptyset = B\cap B'$ fixes $x$. Case 3: Suppose $B = A\backslash B'$ and $|A|= 2$. So $B= \{b\}$ and $B'=\{b'\}$. So $m$ is either the identity function on $A$, the constant function $k$ with image $B$, the constant function $k'$ with image $B'$, or the transposition $(bb')$. Any of the first three functions either maps every member of $B$ to itself or every member of $B'$ to itself, so $mx=x$ if $m$ is one of them; likewise if $m=(bb')$, since $(bb')x = (bb')kx = ((bb')\circ k)x = k'x=x$.