$(\sin x+\cos x)=\sqrt{2}\rightarrow 1+2\sin x\cos x=2\rightarrow 2\sin x\cos x=1\rightarrow \sin x\cos x=\frac{1}{2}$
$a) A=\sin ^{4}x+cos^{4}x=(sin^{2}x+cos^{2}x)^{2}-2sin^{2}xcos^{2}x=1-2.\frac{1}{4}=\frac{1}{2}$
$b) B=(\sin x+\cos x)(sin^{2}x-\sin x\cos x+cos^{2}x)=\sqrt{2}(1-\frac{1}{2})=\frac{1}{\sqrt{2}}$
$c) C=\sin xsin^{4}x +\cos xcos^{4}x=\sin x(\frac{1}{2}-cos^{4}x)+\cos x(\frac{1}{2}-sin^{4}x)=\frac{1}{2}(\sin x+\cos x)-\sin x\cos x(sin^{3}x+cos^{3}x)=\frac{1}{2}.\sqrt{2}-\frac{1}{2}.\frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}$