6. Given four points $A, B, C, D$ on a sphere with radius 3. If $AB=3, CD=4$, then the maximum volume of tetrahedron $ABCD$ is $\qquad$ .
6. $2 \sqrt{5}+3 \sqrt{3}$.

Take the common perpendicular segment $M N$ of the skew lines $A B$ and $C D$, and let the angle between the skew lines $A B$ and $C D$ be $\theta \in\left(0, \frac{\pi}{2}\right]$.
Then $V_{\text {tetrahedron } A B C D}=\frac{1}{6} A B \cdot C D \cdot M N \sin \theta \leqslant 2 M N$.
Let the center of the circumscribed sphere of the tetrahedron $A B C D$ be $O$, and the midpoints of $A B$ and $C D$ be $E$ and $F$, respectively.
Then $O E=\frac{3 \sqrt{3}}{2}, O F=\sqrt{5}$.
The distance between the skew lines $A B$ and $C D$ is
$M N \leqslant E F \leqslant O E+O F=\frac{3 \sqrt{3}}{2}+\sqrt{5}$.
Therefore, $V_{\text {tetrahedron } A B C D} \leqslant 2 M N \leqslant 2 \sqrt{5}+3 \sqrt{3}$.
When $A B \perp C D$, the plane of the small circle with $A B$ as its diameter is parallel to the plane of the small circle with $C D$ as its diameter (the center of the sphere is between the two small circle planes), the equality holds.