= Solution
Write $\Delta M_n=M_n-M_{n-1}$. Conditional on $\mathcal F_{n-1}$, the variables $X_{n-1}$, $M_{n-1}$, and $A_n$ are known, while $\mathbb E[\Delta M_n\mid\mathcal F_{n-1}]=0$. Hence
$$
\begin{aligned}
\operatorname{Cov}(X_n,M_n\mid\mathcal F_{n-1})
&=\operatorname{Cov}(A_n\Delta M_n,\Delta M_n\mid\mathcal F_{n-1})\\
&=A_n\operatorname{Var}(\Delta M_n\mid\mathcal F_{n-1})\\
&=A_n\operatorname{Var}(M_n\mid\mathcal F_{n-1}).
\end{aligned}
$$
The denominator is positive almost surely, so the <recovery of a martingale-transform integrand by conditional covariance> gives
$$
\boxed{A_n=
\frac{\operatorname{Cov}(X_n,M_n\mid\mathcal F_{n-1})}
{\operatorname{Var}(M_n\mid\mathcal F_{n-1})}.}
$$
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