= Solution
First derive the intersection constraints. Since $D$ is <nef> and $F,M$ are effective, $D\cdot F,D\cdot M\geq0$; their sum is $D^2=0$, so both vanish. Since the <movable divisor> $M$ is nef, $M^2,M\cdot F\geq0$, and $M\cdot D=0$ forces both to vanish. In particular
$$
D\cdot F=D\cdot M=M^2=M\cdot F=F^2=0.
$$
For every component $E$ of $F$, nefness and $D\cdot F=0$ imply $D\cdot E=0$. The <isotropic orthogonality consequence of the Hodge index theorem> gives $E^2\leq0$. If $E^2=0$, <Riemann–Roch theorem for algebraic surfaces> and <Serre duality> give $h^0(X,E)\geq2$, since $-E$ cannot be effective. Choose $E'\in|E|$ different from $E$. It cannot contain $E$: otherwise $E'-E$ would be a nonzero effective numerically trivial divisor, contradicting its positive intersection with an ample divisor. Choose a member of $|M|$ avoiding $E$. Replacing one copy of $E$ in $F$ by $E'$ now produces a member of $|D|$ with smaller multiplicity along $E$, contradicting the definition of the <fixed part>. Therefore
$$
\boxed{E^2<0\quad\text{for every component of }F.}
$$
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