a. $A=\frac{4+cos^2x}{3-sin^2x}=\frac{4(1+tan^2x)+1}{3(1+tan^2x)-tan^2x}=\frac{5+4.9}{3+2.9}=\frac{41}{21}$b. $B=(sin^2x+cos^2x)^2-2sin^2x.cos^2x=1-2sin^2x.cos^2x=(1+tan^2x)-2tan^2x=1-3^2=-8$
c. $C=(sin^2x-cos^2x)(sin^4x+sin^2x.cos^2x+cos^4x)=(sin^2x-cos^2x)[(sin^2x+cos^2x)^2-sin^2x.cos^2x]=(sin^2x-cos^2x)(1-sin^2x.cos^2x)=tan^2x-1=3^2-1=8$