a. \(B = \frac{{\cos x + \cos 3x + \cos 5x + \cos 7x}}{{\sin x + \sin 3x + \sin 5x + \sin 7x}}\)

\(\begin{array}{l}
 = \frac{{(\cos x + \cos 7x) + (\cos 3x + \cos 5x)}}{{(\sin x + \sin 7x) + (\sin 3x + \sin 5x)}} = \frac{{2\cos 4x\cos 3x + 2\cos 4x\cos x}}{{2\sin 4x\cos 3x + 2\sin 4x\cos x}} = \\
 = \frac{{\cos 4x}}{{\sin 5x}} = \cot 4x
\end{array}\)

b. Ta có : \(\tan \left( {x + \frac{{7\pi }}{6}} \right) = \tan \left( {x - \frac{\pi }{3} + \frac{{3\pi }}{2}} \right) =  - \cot \left( {x - \frac{\pi }{3}} \right)\)

Do đó : \(\tan \left( {x - \frac{\pi }{3}} \right).\tan \left( {x + \frac{{7\pi }}{6}} \right) =  - 1\)

\(\begin{array}{l}
{\sin ^2}x + {\sin ^2}\left( {\frac{{5\pi }}{3} + x} \right) + {\sin ^2}\left( {\frac{{5\pi }}{3} - x} \right) = \\
 = \frac{{1 - \cos 2x}}{2} + \frac{{1 - \cos (10\pi /3 + 2x)}}{2} + \frac{{1 - \cos (10\pi /3 - 2x)}}{2} = \\
 = \frac{3}{2} - \frac{{\cos 2x + 2\cos (10\pi /3)\cos 2x}}{2} = \frac{3}{2}
\end{array}\)

Suy ra \(C = \frac{3}{2} + 4 = \frac{9}{2}\)