5. Given four points $A, B, C, D$ on a unit sphere, satisfying $AB=AC=AD, BC=CD=DB$. Then the maximum volume of tetrahedron $ABCD$ is $\qquad$

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5. $\frac{8 \sqrt{3}}{27}$.

If the center of the sphere is outside the tetrahedron, there exists a triangle $\triangle B^{\prime} C^{\prime} D^{\prime}$ on the sphere that is parallel to and congruent to $\triangle B C D$, such that the center of the sphere is inside the tetrahedron $A B^{\prime} C^{\prime} D^{\prime}$. Thus, the volume of the tetrahedron $A B^{\prime} C^{\prime} D^{\prime}$ is larger. To find the maximum volume, we only need to consider the case where the center of the sphere is inside the tetrahedron.
As shown in Figure 3, draw the height $A H$, with the center of the sphere $O$ on $A H$.

Let $h=A H, r=H B=H C=H D$. Then
$B C=\sqrt{3} B H=\sqrt{3} r, h \in(1,2)$.
Thus, $V=\frac{1}{3}\left(\frac{\sqrt{3}}{4} B C^{2}\right) A H$
$$
=\frac{1}{3}\left(\frac{\sqrt{3}}{4} \cdot 3 r^{2}\right) h=\frac{\sqrt{3} r^{2} h}{4} \text {. }
$$

By $h=A O+O H=1+\sqrt{1-r^{2}}$
$\Rightarrow(h-1)^{2}=1-r^{2}$
$\Rightarrow h(2-h)=r^{2}$.
Then $V=\frac{\sqrt{3}}{4} h(2-h) h$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{8}(h \cdot h(4-2 h)) \\
\leqslant \frac{\sqrt{3}}{8}\left(\frac{4}{3}\right)^{3}=\frac{8 \sqrt{3}}{27} .
\end{array}
$$

When $h=\frac{4}{3}$, the equality holds.