$$
\begin{aligned}
& \text { Let point } P(a, b) \\
& \qquad \begin{array}{l}
S_1(a, b)=S_2(a, b) \\
\Rightarrow a^2+b^2+2 g a+2 f b+c=a^2+b^2+\frac{3}{2} a+4 b+c=0 \\
a\left(2 g-\frac{3}{2}\right)+b(2 f-4)=0
\end{array}
\end{aligned}
$$
this is the locus of radical axis.
So, $x\left(2 g-\frac{3}{2}\right)+y(2 f-4)=0$ is radical axis of given circles.
This touched the $x^2+y^2+2 x+2 y+1=0$
So, radius $=\sqrt{1^2+1^2-1}=1$ and centre $=(-1,-1)$
So, radius of circle $=$ distance between centre and touching point.
$$
1=\frac{\left|\left(\frac{3}{2}-2 g\right)+(2 f-4)\right|}{\sqrt{\left(\frac{3}{2}-2 g\right)^2+(2 f-4)^2}}
$$
Taking square both sides,
$$
2\left(\frac{3}{2}-2 g\right)(2 f-4)=0
$$
So,
$$
\begin{aligned}
\frac{3}{2}-2 g & =0 \text { or } \\
2 f-4 & =0 \\
2 g & =\frac{3}{2} \text { or } 2 f=4 \\
g & =\frac{3}{4} \text { or } f=2
\end{aligned}
$$