(5) Given a regular tetrahedron $P-ABC$ with the side length of the equilateral triangle base being $2 \sqrt{3}$, and the radius of the inscribed sphere being $\sqrt{2}-1$, then the volume of the tetrahedron is $\qquad$ .
(5) $\sqrt{3}$ Hint: Let the center of the inscribed sphere be $O$, and $D$ be the midpoint of $B C$. Draw $P H \perp$ plane $A B C$ at $H$, then point $O$ is on $P H$.

From the conditions, $H D=\frac{1}{3} \times \frac{\sqrt{3}}{2} \times 2 \sqrt{3}=1, O H=\sqrt{2}-1, \angle P D H=$ $2 \angle O D H$.

Therefore, $\tan \angle O D H=\sqrt{2}-1, \tan \angle P D H=\frac{2(\sqrt{2}-1)}{1-(\sqrt{2}-1)^{2}}=1$, $P H=H D$.
Thus, the volume of the tetrahedron $V=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times(2 \sqrt{3})^{2} \times 1=\sqrt{3}$.