8. As shown in Figure 3, given a regular tetrahedron $O-ABC$ with three lateral edges $OA, OB, OC$ mutually perpendicular, and each of length $2, E, F$ are the midpoints of edges $AB, AC$ respectively, $H$ is the midpoint of line segment $EF$, and a plane is constructed through $EF$ intersecting the lateral edges $OA, OB, OC$ or their extensions at points $A_{1}, B_{1}, C_{1}$. If $OA_{1}=\frac{3}{2}$, then the tangent value of the dihedral angle $O-A_{1}B_{1}-C_{1}$ is $\qquad$
8. $\sqrt{5}$.

As shown in Figure 5, draw $O N \perp A_{1} B_{1}$ at point $N$, and connect $C_{1} N$.
Since $O C_{1} \perp$ plane $O A_{1} B_{1}$, by the theorem of three perpendiculars, we know $C_{1} N \perp A_{1} B_{1}$:
Therefore, $\angle O N C_{1}$ is the plane angle of the dihedral angle $O-A_{1} B_{1}-C_{1}$. Draw $E M \perp O B_{1}$ at point $M$. Then $E M / / O A$.
Thus, $M$ is the midpoint of $O B$. Therefore,
$E M = O M = 1$.
Let $O B_{1} = x$.
From $\frac{O B_{1}}{M B_{1}} = \frac{O A_{1}}{E M} \Rightarrow \frac{x}{x-1} = \frac{3}{2} \Rightarrow x = 3$.
In the right triangle $\triangle O A_{1} B_{1}$, we have
$A_{1} B_{1} = \sqrt{O A_{1}^{2} + O B_{1}^{2}} = \frac{3 \sqrt{5}}{2}$.
Thus, $O N = \frac{O A_{1} \cdot O B_{1}}{A_{1} B_{1}} = \frac{3}{\sqrt{5}}$.
Therefore, $\tan \angle O N C_{1} = \frac{O C_{1}}{O N} = \sqrt{5}$.