8. $E$ and $F$ are the midpoints of edges $BD$ and $CD$ of the regular tetrahedron $ABCD$, respectively. The cosine value of the dihedral angle formed by plane $ABC$ and plane $AEF$ is $\qquad$
8. $\frac{5 \sqrt{33}}{33}$.

Let the edge length of a regular tetrahedron be 1, and the dihedral angle between plane $ABC$ and plane $AEF$ be $\theta$.
It is easy to know that $EF \parallel BC$.
Draw a line $AP$ through point $A$ parallel to $BC$.
Then $EF \parallel BC \parallel AP$, and $AP$ is the edge of the dihedral angle formed by plane $ABC$ and plane $AEF$.
$$
\begin{array}{l}
\text { It is easy to know that } \angle PAC = \angle ACB = 60^{\circ}, \\
\angle CAF = 30^{\circ}, \angle PAF = \angle AFE, \\
AE = AF = \frac{\sqrt{3}}{2}, EF = \frac{1}{2} BC = \frac{1}{2}. \\
\text { Therefore, } \cos \angle AFE = \frac{1}{2 \sqrt{3}}.
\end{array}
$$

By the cosine theorem of trihedral angles, we get
$$
\begin{array}{l} 
\cos \angle CAF \\
= \cos \angle PAC \cdot \cos \angle PAF + \\
\sin \angle PAC \cdot \sin \angle PAF \cdot \cos \theta \\
\Rightarrow \frac{\sqrt{3}}{2} = \frac{1}{2} \times \frac{1}{2 \sqrt{3}} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{11}}{2 \sqrt{3}} \cos \theta \\
\Rightarrow \cos \theta = \frac{5 \sqrt{33}}{33}.
\end{array}
$$