Two circles given are $x^2+y^2+4x-6y-3=0 \& x^2+y^2+4x-2y+1=0$

Now compare $\left(1\right)$ with $x^2+y^2+2gx+2hy+c=0$ we get

$2g=4\Rightarrow g=2 \& 2h=-6 \Rightarrow h=-3$

Centre of circle is given by $\left(-g,-h\right)$ ,so $C_1\left(-2,3\right)$ gives the centre of first circle

Similarly , $C_2\left(-2,1\right)$ which gives centre of second circle.

Now, $r_1=\sqrt{4+9+3}=4 \& r_2=\sqrt{4+1-1}=2$

$C_1{C}_2=2$ (by using distance formula)

$\left|r_1-r_2\right|=C_1{C}_2$

This is the case when a circle is touching another circle internally. So, there will be only one common tangent.