1) Ta có: $A=cos(135^0+360^0.12)-cos(225^0+360^0.2)+tan(-45^0+360^0.3)-cot(-60^0-360^0.4)=cos135^0-cos225^0+tan(-45^0)-cot(-60^0)=-cos45^0+cos45^0-tan45^0+cot60^0=\frac{-3+\sqrt{3}}{3}.$$B=(tan9^0+tan81^0) -(tan27^0+tan63^0)=(tan9^0+cot9^0)-(tan27^0+cot27^0)=(\frac{sin9^0}{cos9^0}+\frac{cos9^0}{sin9^0})-(\frac{sin27^0}{cos27^0}+\frac{cos27^0}{sin27^0})=\frac{sin^29^0+cos^29^0}{sin9^0.cos9^0}-\frac{sin^227^0+cos^227^0}{sin27^0.cos27^0}=\frac{1}{sin9^0.cos9^0}-\frac{1}{sin27^0.cos27^0}=2(\frac{1}{sin18^0}-\frac{1}{sin54^0})=2(\frac{sin54^0-sin18^0}{sin18^0.sin54^0})=4\frac{cos36^0.sin18^0}{sin18^0.sin54^0}=4.\frac{sin54^0}{sin54^0}=4.$