Answer
No solution
Work Step by Step
$$\frac{3}{x-2}+\frac{1}{x+2}=\frac{12}{(x+2)(x-2)}$$
$$((x+2)(x-2))(\frac{3}{x-2}+\frac{1}{x+2})=((x+2)(x-2))(\frac{12}{(x+2)(x-2)})$$
$$3(x+2)+(x-2)=12$$
$$3x+6+x-2=12$$
$$4x+4=12$$
$$4x=8$$
$$x=2$$
2 is an extraneous solution; there is no solution.