$\begin{array}{l} {\sin ^4}x + {\cos ^4}x = {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2} + {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)^2}\\ = \dfrac{{1 - 2\cos 2x + {{\cos }^2}2x + 1 + 2\cos 2x + {{\cos }^2}2x}}{4}\\ = \dfrac{{2{{\cos }^2}2x + 2}}{4} = \dfrac{{2\left( {\dfrac{{1 + \cos 4x}}{2}} \right) + 2}}{4} = \dfrac{{3 + \cos 4x}}{4} = \dfrac{1}{4}\left( {3 + \cos 4x} \right)\\ b){\sin ^6}x + {\cos ^6}x = \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)\\ = {\sin ^4}x - \dfrac{1}{4}{\sin ^2}2x + {\cos ^4}x\\ = {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2} - \dfrac{1}{4}\left( {\dfrac{{1 - \cos 4x}}{2}} \right) + {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2}\\ = \dfrac{1}{4}\left( {3 + \cos 4x} \right) - \dfrac{1}{4}\left( {\dfrac{1}{2} - \dfrac{{\cos 4x}}{2}} \right) = \dfrac{5}{8} + \dfrac{3}{8}\cos 4x\\ c)\cos 3x{\sin ^3}x + \sin 3x{\cos ^3}x = \cos 3x.\dfrac{1}{4}\left( {3\sin x - \sin 3x} \right) + \dfrac{1}{4}\left( {3\cos x + \cos 3x} \right)\sin 3x\\ = \dfrac{1}{4}\cos 3x.3\sin x - \dfrac{1}{4}\cos 3x.\sin 3x + \dfrac{1}{4}.3\cos x.\sin 3x + \dfrac{1}{4}\cos 3x\sin 3x\\ = \dfrac{3}{4}\left( {\cos 3x.\sin x + \sin 3x.\cos x} \right) = \dfrac{3}{4}\cos 4x\\ d)\tan x\left( {\dfrac{{1 + {{\cos }^2}x}}{{\sin x}} - \sin x} \right) = \tan x\left( {\dfrac{{1 + {{\cos }^2}x - {{\sin }^2}x}}{{\sin x}}} \right) = \dfrac{{\sin x}}{{\cos x}}.\dfrac{{{{\cos }^2}x + {{\cos }^2}x}}{{\sin x}} = 2\cos x \end{array}$
