$a)\alpha=30^o\\ A=\dfrac{\cos\alpha+\sin 2\alpha}{2\cos2\alpha-\sin\alpha}\\ =\dfrac{\cos(30^o)+\sin(60^o)}{2\cos(60^o)-\sin(30^o)}\\ =\dfrac{\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}}{2.\dfrac{1}{2}-\dfrac{1}{2}}\\ =\dfrac{\sqrt{3}}{\dfrac{1}{2}}\\ =2\sqrt{3}\\ b)\alpha=45^o\\ B=\dfrac{\cos\alpha+\sin 2\alpha}{2\cos2\alpha-\sin\alpha}+2\tan\alpha\\ =\dfrac{\cos(45^o)+\sin(90^o)}{2\cos(90^o)-\sin(45^o)}+2\tan(45^o)\\ =\dfrac{\dfrac{\sqrt{2}}{2}+1}{2.0-\dfrac{\sqrt{2}}{2}}+2.1\\ =\dfrac{\dfrac{2+\sqrt{2}}{2}}{-\dfrac{\sqrt{2}}{2}}+2\\ =1-\sqrt{2}\\ c)\alpha=15^o\\ C=\dfrac{\cos 3\alpha-\sin 2\alpha}{2\cos 2\alpha-\sin 3\alpha}-\cot 4\alpha\\ =\dfrac{\cos(45^o)-\sin(30^o)}{2\cos(30^o)-\sin(45^o)}-\cot(60^o)\\ =\dfrac{\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}}{2.\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}}-\dfrac{\sqrt{3}}{3}\\ =\dfrac{\dfrac{-1+\sqrt{2}}{2}}{\dfrac{2\sqrt{3}-\sqrt{2}}{2}}-\dfrac{\sqrt{3}}{3}\\ =\dfrac{-1+\sqrt{2}}{2\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}}{3}\\ =\dfrac{(-1+\sqrt{2})(2\sqrt{3}-\sqrt{2})}{(2\sqrt{3}-\sqrt{2})(2\sqrt{3}+\sqrt{2})}-\dfrac{\sqrt{3}}{3}\\ =\dfrac{-2\sqrt{3}+\sqrt{2}-2\sqrt{6}-2}{10}-\dfrac{\sqrt{3}}{3}\\ =\dfrac{-6\sqrt{3}+3\sqrt{2}-6\sqrt{6}-6-10\sqrt{3}}{30}\\ =\dfrac{3\sqrt{2}-6\sqrt{6}-6-16\sqrt{3}}{30}\\ d)\alpha=20^o\\ D=\dfrac{\cos 3\alpha+\sin 3\alpha}{2\cos 3\alpha-\tan 3\alpha}\\ =\dfrac{\cos(60^o)+\sin(60^o)}{2\cos(60^o)-\tan(60^o)}\\ =\dfrac{\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}}{2.\dfrac{1}{2}-\sqrt{3}}\\ =\dfrac{\dfrac{1+\sqrt{3}}{2}}{1-\sqrt{3}}\\ =\dfrac{(1+\sqrt{3})^2}{2(1-\sqrt{3})(1+\sqrt{3})}\\ =\dfrac{4+2\sqrt{3}}{-4}\\ =-\dfrac{2+\sqrt{3}}{2}$