Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\dfrac{\pi }{2} < a < \pi  \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
\cos a < 0 \Rightarrow \cos a =  - \sqrt {1 - {{\sin }^2}a}  =  - \dfrac{{\sqrt 6 }}{3}\\
\sin \left( {a + \dfrac{\pi }{6}} \right) = \sin a.\cos \dfrac{\pi }{6} + \cos a.\sin \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}.\dfrac{{\sqrt 3 }}{2} + \left( { - \dfrac{{\sqrt 6 }}{3}} \right).\dfrac{1}{2} = \dfrac{{3 - \sqrt 6 }}{6}\\
b,\\
\dfrac{{3\pi }}{2} < a < 2\pi  \Rightarrow \left\{ \begin{array}{l}
\sin a < 0\\
\cos a > 0
\end{array} \right.\\
\sin a < 0 \Rightarrow \sin a =  - \sqrt {1 - {{\cos }^2}a}  =  - \dfrac{{\sqrt 6 }}{3}\\
\cos \left( {a + \dfrac{\pi }{3}} \right) = \cos a.\cos \dfrac{\pi }{3} - \sin a.\sin \dfrac{\pi }{3} = \dfrac{1}{{\sqrt 3 }}.\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}.\left( { - \dfrac{{\sqrt 6 }}{3}} \right) = \dfrac{{\sqrt 3  + 3\sqrt 2 }}{6}\\
c,\\
P = \sin \dfrac{\pi }{{24}}.\cos \dfrac{\pi }{6}.\cos \dfrac{\pi }{{12}}.\cos \dfrac{\pi }{{24}}\\
 = \dfrac{1}{2}.\left( {2\sin \dfrac{\pi }{{24}}.\cos \dfrac{\pi }{{24}}} \right).\cos \dfrac{\pi }{{12}}.\cos \dfrac{\pi }{6}\\
 = \dfrac{1}{2}.sin\dfrac{\pi }{{12}}.\cos \dfrac{\pi }{{12}}.\cos \dfrac{\pi }{6}\\
 = \dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{\pi }{6}.\cos \dfrac{\pi }{6}\\
 = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{\pi }{3}\\
 = \dfrac{1}{8}.\dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt 3 }}{{16}}\\
d,\\
\cos x + \cos y = 2\cos \dfrac{{x + y}}{2}.\cos \dfrac{{x - y}}{2}\\
S = \cos \dfrac{{2\pi }}{9} + \cos \dfrac{{4\pi }}{9} + \cos \dfrac{{8\pi }}{9}\\
 = 2.\cos \dfrac{{3\pi }}{9}.\cos \dfrac{\pi }{9} - \cos \left( {\pi  - \dfrac{{8\pi }}{9}} \right)\\
 = 2.cos\dfrac{\pi }{3}.\cos \dfrac{\pi }{9} - \cos \dfrac{\pi }{9}\\
 = \cos \dfrac{\pi }{9}\left( {2\cos \dfrac{\pi }{3} - 1} \right)\\
 = \cos \dfrac{\pi }{9}.0 = 0
\end{array}\)