2. Let $\triangle A B C$ have interior angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and satisfy
$$
\begin{array}{l}
a \cos B-b \cos A=\frac{3}{5} c . \\
\text { Then } \frac{\tan A}{\tan B}=
\end{array}
$$
2.4.

Solution 1 From the given and the cosine rule, we have
$$
\begin{array}{l}
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \\
\Rightarrow a^{2}-b^{2}=\frac{3}{5} c^{2} . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}} \\
=\frac{c^{2}+a^{2}-b^{2}}{c^{2}+b^{2}-a^{2}}=4 .
\end{array}
$$

Solution 2 Draw $C D \perp A B$ from point $C$, with the foot of the perpendicular at $D$. Then
$a \cos B=D B$,
$b \cos A=A D$.
From the given, we have
$D B-A D=\frac{3}{5} c$.
Figure 4
Also, $D B+D A=c$,

Solving these equations, we get
$$
\begin{array}{l}
A D=\frac{1}{5} c, D B=\frac{4}{5} c . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\frac{C D}{A D}}{\frac{C D}{D B}}=\frac{D B}{A D}=4 .
\end{array}
$$

Solution 3 From the projection theorem, we have $a \cos B+b \cos A=c$.
Also, $a \cos B-b \cos A=\frac{3}{5} c$. Solving these equations, we get $a \cos B=\frac{4}{5} c, b \cos A=\frac{1}{5} c$.
Therefore, $\frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{a \cos B}{b \cos A}=4$.