19. As shown in the figure, in the dihedral angle $\alpha-E F-\beta$, $A E \subset \alpha, B F \subset \beta$, and $A E \perp E F, B F \perp E F, E F=1, A E=2, A B=$ $\sqrt{2}$, find the maximum volume of the tetrahedron $A B E F$.
19. As shown in the figure, construct $CE$ parallel to $BF$, and connect $AC, BC$. Since $BCFE$ is a rectangle, we have $V_{ABEF}=V_{ABBC}=V_{B-AEC}$. Given that $EF \perp AE, EF \perp EC$, it follows that $EF \perp$ plane $AEC$.
$\therefore BC \perp$ plane $AEC$.
$\therefore BC=EF=1, AC=\sqrt{AB^{2}-BC^{2}}=1$.
Since $BC$ is the height of the tetrahedron $B-AEC$, to maximize $V_{BAEC}$, we only need to maximize $S_{\triangle ABC}$. Clearly, when $\angle EAC=90^{\circ}$, $\left(S_{\triangle ABC}\right)=\frac{1}{2} \cdot AE \cdot AC=\frac{1}{2} \cdot 2 \cdot 1=1$.
$\therefore$ The maximum volume of the tetrahedron $ABEF$ is $\frac{1}{3}$.