2. In a right-angled tetrahedron $ABCD$, if the sum of the six edge lengths is 6, then the maximum value of its volume is $\qquad$ .
2. $\frac{4}{3(1+\sqrt{2})^{3}}$.

Let edges $D A$, $D B$, and $D C$ be mutually perpendicular.
Let $D A=a$, $D B=b$, $D C=c$.
Then $V=\frac{1}{6} a b c \Rightarrow a b c=6 V$.
And $A B=\sqrt{a^{2}+b^{2}}$, $B C=\sqrt{b^{2}+c^{2}}$,
$C A=\sqrt{c^{2}+a^{2}}$,
Thus $6=a+b+c+\sqrt{a^{2}+b^{2}}+$
$$
\begin{aligned}
& \sqrt{b^{2}+c^{2}}+\sqrt{c^{2}+a^{2}} \\
\geqslant & 3 \sqrt[3]{a b c}+\sqrt{2 a b}+\sqrt{2 b c}+\sqrt{2 c a} \\
\geqslant & 3 \sqrt[3]{a b c}+3 \sqrt[3]{2 \sqrt{2} a b c} \\
= & 3 \sqrt[3]{6}(1+\sqrt{2}) \sqrt[3]{V} \\
\Rightarrow & V \leqslant \frac{4}{3(1+\sqrt{2})^{3}} .
\end{aligned}
$$

When $a=b=c$, the equality holds.