Let y=f(x)=(x+2)/(x-2), f(3)=(3+2)/(3-2)=5.
y(x-2)=(x+2), xy-2y=x+2, xy-x=2y+2, x(y-1)=2y+2, x=(2y+2)/(y-1), and x, by definition, equals f⁻¹(y)=(2y+2)/(y-1). So f⁻¹(2)=(4+2)/(2-1)=6.
Now let y=g(x)=mx+c, mx=y-c, x=g⁻¹(y)=(y-c)/m. g⁻¹(2)=(2-c)/m=f(3)=5. So 2-c=5m. 5m+c=2.
If gf⁻¹(2) is meant to be g(f⁻¹(2))=g(6)=m(6)+c=6m+c, so 6m+c=5 and c=5-6m. 5m+c=2 becomes 5m+5-6m=2. From this 5-2=6m-5m=m, so m=3. Therefore c=5-18=-13. g(x)=3m-13. m=3 and c=-13.