We have,
$$
\begin{aligned}
x^2+y^2-6 x+4 y & =12 \\
\Rightarrow \quad(x-3)^2+(y+2)^2 & =25 \\
\Rightarrow \quad(x-3)^2+(y+2)^2 & =(5)^2
\end{aligned}
$$
Equation of tangent whose slope $m$ is
$$
y+2=m(x-3) \pm 5 \sqrt{m^2+1}
$$
Now, this tangent is parallel to line $4 x+3 y+5=0$
$\therefore$ Slope of line is $-\frac{4}{3}$
Put the value of $m=-\frac{4}{3}$ is Eq. (i), we get
$$
y+2=-\frac{4}{3}(x-3) \pm 5 \sqrt{\left(\frac{-4}{3}\right)^2+1}
$$

$$
\begin{aligned}
& \Rightarrow \quad y+2=\frac{-4}{3}(x-3) \pm 5\left(\frac{5}{3}\right) \\
& \Rightarrow \quad 3 y+6=-4 x+12 \pm 25 \\
& \Rightarrow 4 x+3 y=6 \pm 25 \\
& \Rightarrow 4 x+3 y=31 \text { or } 4 x+3 y=-19
\end{aligned}
$$
Hence, equation of tangent is
$$
4 x+3 y-31=0 \text { or } 4 x+3 y+19=0
$$