tan a = 8/15 .

Vì \(a,b\) là các góc nhọn nên \(\cos a > 0,\cos b > 0\).

Ta có \(\cos a = \sqrt {1 - {{\sin }^2}a} = \frac{{15}}{{17}} \Rightarrow \tan a = \frac{{\sin a}}{{\cos a}} = \frac{8}{{15}}\);

            \(\cos b = \sqrt {\frac{1}{{1 + {{\tan }^2}b}}} = \frac{{12}}{{13}} \Rightarrow \sin b = \cos b\tan b = \frac{5}{{13}}{\rm{. }}\)

Khi đó, \(\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b = \frac{8}{{17}} \cdot \frac{{12}}{{13}} - \frac{{15}}{{17}} \cdot \frac{5}{{13}} = \frac{{21}}{{221}}\).

            \(\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b = \frac{{15}}{{17}} \cdot \frac{{12}}{{13}} - \frac{8}{{17}} \cdot \frac{5}{{13}} = \frac{{140}}{{221}}\)

            \(\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{\frac{8}{{15}} + \frac{5}{{12}}}}{{1 - \frac{8}{{15}} \cdot \frac{5}{{12}}}} = \frac{{171}}{{140}}.\)

Đáp án:           a) Đúng,          b) Đúng,         c) Sai,              d) Sai.