Solution: 4.
As shown in Figure 4, connect $P Q$, then $P Q \perp$ plane $A B C$, with the foot of the perpendicular $H$ being the center of the equilateral $\triangle A B C$, and $P Q$ passing through the center of the sphere $O$. Connect $C H$ and extend it to intersect $A B$ at point $M$, then $M$ is the midpoint of $A B$, and $C M \perp A B$. It is easy to see that $\angle P M H$ and $\angle Q M H$ are the plane angles of the dihedral angles formed by the lateral faces and the base of the regular triangular pyramids $P-A B C$ and $Q-A B C$, respectively, so $\angle P M H=45^{\circ}$, thus $P H=M H=\frac{1}{2} A H$.
Since $\angle P A Q=90^{\circ}, A H \perp P Q$, we have
$$
A H^{2}=P H \cdot Q H \text {, i.e., } A H^{2}=\frac{1}{2} A H \cdot Q H .
$$

Therefore, $Q H=2 A H=4 M H$.
Hence, $\tan \angle Q M H=\frac{Q H}{M H}=4$.
5. Let two regular tetrahedra $P-A B C$ and $Q-A B C$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-A B C$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-A B C$ is $\qquad$ .