Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\frac{\pi }{2} < a < \pi  \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
\cos a < 0 \Rightarrow \cos a =  - \sqrt {1 - {{\sin }^2}a}  =  - \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}}  =  - \frac{4}{5}\\
\tan \left( {a + \frac{\pi }{3}} \right) = \frac{{\sin \left( {a + \frac{\pi }{3}} \right)}}{{\cos \left( {a + \frac{\pi }{3}} \right)}}\\
 = \frac{{\sin a.cos\frac{\pi }{3} + \sin \frac{\pi }{3}.\cos a}}{{\cos a.sin\frac{\pi }{3} - \sin a.\cos \frac{\pi }{3}}}\\
 = \frac{{\frac{3}{5}.\frac{1}{2} + \frac{{\sqrt 3 }}{2}.\frac{{ - 4}}{5}}}{{\frac{{ - 4}}{5}.\frac{{\sqrt 3 }}{2} - \frac{3}{5}.\frac{1}{2}}}\\
 = \frac{{3 - 4\sqrt 3 }}{{ - 3 - 4\sqrt 3 }} = \frac{{4\sqrt 3  - 3}}{{4\sqrt 3  + 3}}\\
b,\\
\frac{{3\pi }}{2} < a < 2\pi  \Rightarrow \left\{ \begin{array}{l}
\sin a < 0\\
\cos a > 0
\end{array} \right.\\
\tan a = \frac{{ - 12}}{{13}} \Leftrightarrow \frac{{\sin a}}{{\cos a}} =  - \frac{{12}}{{13}} \Leftrightarrow \sin a =  - \frac{{12}}{{13}}\cos a\\
{\sin ^2}a + {\cos ^2}a = 1\\
 \Leftrightarrow {\left( { - \frac{{12}}{{13}}\cos a} \right)^2} + {\cos ^2}a = 1\\
 \Leftrightarrow \frac{{313}}{{169}}{\cos ^2}a = 1\\
 \Leftrightarrow {\cos ^2}a = \frac{{169}}{{313}}\\
\cos a > 0 \Rightarrow \cos a = \frac{{13}}{{\sqrt {313} }}\\
\sin a = \frac{{ - 12}}{{13}}\cos a =  - \frac{{12}}{{\sqrt {313} }}\\
\cos \left( {\frac{\pi }{3} - a} \right) = \cos \frac{\pi }{3}.\cos a + \sin \frac{\pi }{3}.\sin a = \frac{1}{2}.\frac{{ - 12}}{{\sqrt {313} }} + \frac{{\sqrt 3 }}{2}.\frac{{13}}{{\sqrt {313} }} = \frac{{13\sqrt 3  - 12}}{{2\sqrt {313} }}
\end{array}\)