8. Given a regular tetrahedron $P-ABC$ with a base edge length of 1, and the distance from point $P$ to the base $ABC$ is $\sqrt{2}$. Then the radius of the inscribed sphere of the tetrahedron is $\qquad$ .
8. $\frac{\sqrt{2}}{6}$.

As shown in Figure 1, draw $P O_{1} \perp$ plane $A B C$ at point $O_{1}$. Take the midpoint $D$ of $A B$, and connect $C D$.
Then $O_{1} D \perp A B$.
Connect $P D$.
Then $P D \perp A B$.
According to the problem, the incenter of the regular tetrahedron $P-A B C$ lies on the line segment $P O_{1}$, denoted as $O$.

In the plane $P O_{1} D$, draw $O F \perp P D$, with the foot of the perpendicular at $F$. Then $O F$ is the inradius $r$ of the regular tetrahedron $P-A B C$.
$$
\begin{array}{l}
\text { Clearly, } O O_{1}=r . \\
\text { In Rt } \triangle P O_{1} D \text {, } \\
\text { Rt } \triangle P O_{1} D \backsim \mathrm{Rt} \triangle P F O \Rightarrow \frac{P D}{O_{1} D}=\frac{P O}{F O} \\
\Rightarrow \frac{\sqrt{P O_{1}^{2}+O_{1} D^{2}}}{\frac{1}{3} C D}=\frac{\sqrt{2}-r}{r} \Rightarrow r=\frac{\sqrt{2}}{6} .
\end{array}
$$