Ответ:

  1. \(\frac{\sin^2 315^\circ\cdot\cos300^\circ+ g(-315^\circ)}{\sin(-120^\circ)\cdot\cos150^\circ}=\frac{\frac12\cdot\frac12+1}{(-\frac{\sqrt3}{2})(-\frac{\sqrt3}{2})}=\frac{\frac54}{\frac34}=\frac53\).
  2. \(\frac{6\cos^2(-240^\circ)\cdot\ctg210^\circ}{\sin(-300^\circ)\cdot\cos^2 180^\circ}=\frac{6\cdot\frac14\cdot\sqrt3}{\frac{\sqrt3}{2}\cdot1}=\frac{3\sqrt3}{2}\cdot\frac{2}{\sqrt3}=3\).

Ответ: 1) \(\frac53\); 2) 3.