5. As shown in Figure 1, given $\angle A C B=90^{\circ}, D A \perp$ plane $A B C, A E \perp D B$ at point $E, A F \perp D C$ at point $F$, and $A D=A B=2$. Then the maximum volume of the tetrahedron $D-A E F$ is . $\qquad$
5. $\frac{\sqrt{2}}{6}$.

First, prove: $D E \perp$ plane $A E F$.
Given $A D \perp B C, A C \perp B C$
$\Rightarrow B C \perp$ plane $A C D$
$\Rightarrow \angle D C B=90^{\circ}$.
By the projection theorem,
$A D^{2}=D E \cdot D B, A D^{2}=D F \cdot D C$
$\Rightarrow D E \cdot D B=D F \cdot D C$
$\Rightarrow E, B, C, F$ are concyclic
$\Rightarrow E F \perp E D$.
Also, $A E \perp B D$, so $D E \perp$ plane $A E F$.
Next, prove: $\angle A F E=90^{\circ}$.
Given $B C \perp A C, B C \perp A D$
$\Rightarrow B C \perp$ plane $A C D \Rightarrow B C \perp A F$.
Also, $A F \perp C D$, so $A F \perp$ plane $D B C$.
Thus, $A F \perp E F$.
Therefore, $V_{\text {tetrahedron } D-A E F}=\frac{1}{6} D E \cdot A F \cdot E F$
$$
\leqslant \frac{1}{12} D E \cdot A E^{2}=\frac{\sqrt{2}}{6}
$$