8. Let $\triangle A B C$ have the sides opposite to $\angle A, \angle B, \angle C$ be $a, b, c$ respectively, and satisfy $a \cos B - b \cos A = \frac{4}{5} c$. Then $\frac{\tan A}{\tan B}=$ $\qquad$.
8.9.

From the cosine theorem, we have
$$
\begin{array}{l}
a \cdot \frac{a^{2}+c^{2}-b^{2}}{2 a c}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{4}{5} c \\
\Rightarrow a^{2}-b^{2}=\frac{4}{5} c^{2} .
\end{array}
$$

Thus, from the sine theorem, we get
$$
\begin{array}{l}
\frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{a \cdot \frac{a^{2}+c^{2}-b^{2}}{2 a c}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}} \\
=\frac{a^{2}+c^{2}-b^{2}}{b^{2}+c^{2}-a^{2}}=9 .
\end{array}
$$