6. Among all tetrahedra with edge lengths $2,3,3,4,5,5$, the maximum volume is $\qquad$
6. $\frac{8 \sqrt{2}}{3}$.

Consider a triangle with one side of length 2, the other two sides can only have the following four scenarios: (a) 3,3, (b) 5,5, (c) 4,5; (d) 3,4.

Thus, a tetrahedron with a base of a triangle with side length 2 has only three possible configurations.
(1) Base $2,3,3$, side $2,5,5$. In this case, $C D \perp$ base $A B C, h=C D=4, S_{\triangle A B C}=$. $2 \sqrt{2}, V_{1}=\frac{1}{3} S_{\triangle A B C h_{1}}=$ $8 \sqrt{2}$
(2) Base $2,3,3$, side $2,4,5$. Since the height is no greater than the altitude, then $h_{2} \leqslant A D=$ $4, V_{2} \leqslant \frac{8 \sqrt{2}}{3}$.
(3) Base $2,3,4$, side

$2,5,5 . h_{3} \leqslant C D=3, S_{\triangle A B C}=\frac{3}{4} \sqrt{15}$,
$$
\begin{array}{l}
\text { then } V_{3} \leqslant \frac{1}{3} \cdot \frac{3}{4} \sqrt{15} \cdot 3 \\
=\frac{3}{4} \sqrt{15} \\
<\frac{8 \sqrt{2}}{3} .
\end{array}
$$

Thus,
$$
V_{\max }=\frac{8 \sqrt{2}}{3}
$$