Giải thích các bước giải: \(\begin{array}{l} 1.\cos ({795^0}) = \cos ({2.360^0} + {75^0}) = \cos {75^0} = \cos ({30^0} + {45^0}) = \cos {30^0}.cos{45^0} - sin{30^0}.sin{45^0} = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 2 }}{2} - \frac{1}{2}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 - \sqrt 2 }}{4}\\ \tan \left( {\frac{{7\pi }}{{12}}} \right) = \tan \left( {\frac{\pi }{4} + \frac{\pi }{3}} \right) = \frac{{\tan \frac{\pi }{4} + \tan \frac{\pi }{3}}}{{1 - \tan \frac{\pi }{4}.\tan \frac{\pi }{3}}} = \frac{{1 + \sqrt 3 }}{{1 - \sqrt 3 }} = - 2 - \sqrt 3 \\ \cot \left( {\frac{{5\pi }}{8}} \right) = 1 - \sqrt 2 \\ 2.\sin a = \frac{8}{{17}};tanb = \frac{5}{{12}}\\ 0 < a,b < 90 \Rightarrow \cos a > 0;\cos b > 0;\sin b > 0\\ {\cos ^2}a + {\sin ^2}a = 1 \Rightarrow \cos a = \frac{{15}}{{17}} \Rightarrow \tan a = \frac{{\sin a}}{{\cos a}} = \frac{8}{{15}}\\ 1 + {\tan ^2}b = \frac{1}{{{{\cos }^2}b}} \Rightarrow \cos b = \frac{{12}}{{13}} \Rightarrow \sin b = \cos b.\tan b = \frac{5}{{13}}\\ sin(a - b) = \sin a.\cos b - \cos a.\sin b = \frac{8}{{17}}.\frac{{12}}{{13}} - \frac{{15}}{{17}}.\frac{5}{{13}} = \frac{{21}}{{221}}\\ \cos (a - b) = \cos a.\cos b + \sin a.\sin b = \frac{{12}}{{13}}.\frac{{15}}{{17}} + \frac{5}{{13}}.\frac{8}{{17}} = \frac{{220}}{{221}}\\ \tan (a - b) = \frac{{\sin (a - b)}}{{\cos (a - b)}} = \frac{{21}}{{220}}\\ 3.\sin a = \frac{5}{{13}};\frac{\pi }{2} < a < \pi \Rightarrow \cos a < 0\\ {\sin ^2}a + {\cos ^2}a = 1 \Rightarrow \cos a = \frac{{ - 12}}{{13}}\\ \sin b = \frac{3}{5};0 < b < \frac{\pi }{2} \Rightarrow \cos a > 0 \Rightarrow \cos b = \frac{4}{5}\\ \sin (a + b) = \sin a.\cos b + \cos a.\sin b = \frac{{ - 16}}{{65}}\\ 4.\tan a = \frac{1}{7};\tan b = \frac{3}{4}\\ \tan (a + b) = \frac{{\tan a + \tan b}}{{1 - \tan a.\tan b}} = 1 \Rightarrow a + b = 45 + 180k = \frac{\pi }{4} + k\pi (k \in Z) \end{array}\)