The equations of the tangents to the circle $x^2+y^2-4x-6y-12=0$ and parallel to the line $4x-3y=1$ are

For the circle $x^2+y^2-4x-6y-12=0$, centre is $\left(2,3\right),$ radius =5

Equation of a line parallel to 4x-3y=1 is 4x-3y+k=0

4x-3y+k=0 is a tangent to the circle$\Rightarrow$$\frac{\left|4\left(2\right)-3\left(3\right)+k\right|}{5}$=5$\Rightarrow \left|k-1\right|=25\Rightarrow k=1\pm 25$

$\Rightarrow k=26 or -24.$                   $\therefore$ Required tangents are $4x-3y+26=0, 4x-3y-24=0.$