12. In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=30^{\circ}, A C=2, M$ is the midpoint of $A B$. If $\triangle A C M$ is folded along $C M$ such that the distance between $A$ and $B$ is $2 \sqrt{2}$, then the volume of the tetrahedron $A-B C M$ is $\qquad$
Draw $A D \perp B C$, and connect $M D$.
Then $A C=2, B C=2 \sqrt{3} \Rightarrow \angle B A C=90^{\circ}$
$\Rightarrow \cos \angle A B C=\frac{\sqrt{6}}{3} \Rightarrow B D=\frac{4 \sqrt{3}}{3}$
$\Rightarrow \angle B M D=90^{\circ} \Rightarrow B M \perp$ plane $A M D$
$\Rightarrow A D \perp$ plane $B M C$.
Thus, $S_{\triangle B M C}=\sqrt{3}, A D=A B \sin \angle A B C=\frac{2 \sqrt{6}}{3}$,
so the volume of the tetrahedron $A-B C M$ is $V=\frac{1}{3} \cdot \sqrt{3} \cdot \frac{2 \sqrt{6}}{3}=\frac{2 \sqrt{2}}{3}$.