Let $(a, b)$ be the point where line $4 x-3 y+7=0$ touches the circle $x^2+y^2-6 x+4 y-12=0$.
Now, equation of the tangent at $(a, b)$ to the given circle is
$$
\begin{aligned}
x a+y b-3(x+a)+2(y+b)-12 & =0 \\
\Rightarrow \quad(a-3) x+(b+2) y-(3 a-2 b+12) & =0
\end{aligned}
$$
If it represents the given line $4 x-3 y+7=0$
Then, $\frac{a-3}{4}=\frac{b+2}{-3}=\frac{-(3 a-2 b+12)}{7}=1 \quad$ (say)
$$
\begin{array}{lrrl}
\Rightarrow & a=4 l+3, b=-3 l-2 \\
\text { and } & -(3 a-2 b+12)=7 l \\
\Rightarrow & -3 a+2 b-12=7 l \\
\Rightarrow & -3(4 l+3)+2(-3 l-2)-12=7 l \\
\Rightarrow & & l=-1 \\
\therefore & & a=-1 \text { and } b=1
\end{array}
$$