(2) Let the internal angles $A, B, C$ of $\triangle ABC$ have opposite sides $a, b, c$ respectively, and satisfy $a \cos B - b \cos A = \frac{3}{5} c$. Then the value of $\frac{\tan A}{\tan B}$ is $\qquad$.
(2) 4 Hint: Method one From the given and the cosine rule, we get
$$
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c,
$$

which means $a^{2}-b^{2}=\frac{3}{5} c^{2}$.
Therefore, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}}$
$$
=\frac{c^{2}+a^{2}-b^{2}}{b^{2}+c^{2}-a^{2}}=\frac{\frac{8}{5} c^{2}}{\frac{2}{5} c^{2}}=4 \text {. }
$$

Method two As shown in the figure, draw $C D \perp A B$, with the foot of the perpendicular at $D$, then
$$
a \cos B=D B, b \cos A=A D \text {. }
$$

From the given, we have $D B-A D=\frac{3}{5} c$.
Also, $D B+D A=c$, solving these equations simultaneously gives $A D=$ $\frac{1}{5} c, D B=\frac{4}{5} c$.
$$
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\frac{C D}{A D}}{\frac{C D}{D B}}=\frac{D B}{A D}=4 \text {. }
$$

Method three From the projection theorem, we get $a \cos B+b \cos A=c$.
Also, $a \cos B-b \cos A=\frac{3}{5} c$, solving these equations simultaneously gives $a \cos B=\frac{4}{5} c, b \cos A=\frac{1}{5} c$. Therefore, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cos B}{b \cos A}=\frac{\frac{4}{5} c}{\frac{1}{5} c}=4$.