Answer $\frac{\sqrt{2}}{6}$.
Solution As shown in the figure, let the projections of the center of the sphere $O$ on the plane $ABC$ and the plane $ABP$ be $H$ and $K$, respectively, and let the midpoint of $AB$ be $M$, and the radius of the inscribed sphere be $r$. Then $P, K, M$ are collinear, $P, O, H$ are collinear, $\angle PHM = \angle PKO = \frac{\pi}{2}$, and
$$
\begin{array}{c}
OH = OK = r, \quad PO = PH - OH = \sqrt{2} - r, \\
MH = \frac{\sqrt{3}}{6} AB = \frac{\sqrt{3}}{6}, \quad PM = \sqrt{MH^2 + PH^2} = \sqrt{\frac{1}{12} + 2} = \frac{5 \sqrt{3}}{6},
\end{array}
$$

Thus, we have
$$
\frac{r}{\sqrt{2} - r} = \frac{OK}{PO} = \sin \angle KPO = \frac{MH}{PM} = \frac{1}{5},
$$

Solving this, we get $r = \frac{\sqrt{2}}{6}$.
4. Given a regular tetrahedron $P-ABC$ with a base edge length of 1 and a height of $\sqrt{2}$, then its inscribed sphere radius is