jmc

Multiplying both sides by $(x-2)(x+2)(x^2+4),$ we get $$x^3+3x^2-12x+36=A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2).$$Setting $x=2,$ we get $32A=32,$ so $A=1.$

Setting $x=-2,$ we get $-32B=64,$ so $B=-2.$ Then $$x^3+3x^2-12x+36=(x+2)(x^2+4)-2(x-2)(x^2+4)+(Cx+D)(x-2)(x+2).$$This simplifies to $$2x^3-3x^2-8x+12=(Cx+D)(x-2)(x+2),$$which factors as $$(2x-3)(x-2)(x+2)=(Cx+D)(x-2)(x+2).$$Hence, $C=2$ and $D=-3,$ so $(A,B,C,D)=\boxed{(1,-2,2,-3)}.$