5. Let two regular tetrahedra $P-ABC$ and $Q-ABC$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-ABC$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-ABC$ is . $\qquad$
5.4.

As shown in Figure 6, 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 side $A B$, 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 tetrahedrons $P-A B C$ and $Q-A B C$, respectively.
$$
\begin{array}{l}
\text { Then } \angle P M H=45^{\circ} \Rightarrow P H=M H=\frac{1}{2} A H \text {. } \\
\text { By } \angle P A Q=90^{\circ}, A H \perp P Q \\
\Rightarrow A H^{2}=P H \cdot Q H \\
\Rightarrow A H^{2}=\frac{1}{2} A H \cdot Q H \\
\Rightarrow Q H=2 A H=4 M H \text {. } \\
\text { Therefore } \tan \angle Q M H=\frac{Q H}{M H}=4 .
\end{array}
$$