Problem 1
Given $\sin \left(6a\right)=-\frac{\sqrt{5}}{3}$ and $\cos \left(6a\right)>0$, Find $\sin \left(3a\right)$ and $\tan \left(12a\right)$
Problem 2
Given $\sin \left(\frac{1}{2}a\right)=-\frac{2}{3}$ and that $a/2$ is in Quadrant III, Find: $\sin(a) and $\cos (a/4 )$
Progress
Completely stuck on how to do $6a$, I've learned $2a$ but I have no idea how to apply it for 6. Also, how do I deal with the negative/positive signs in each question?
Since you know that $\sin 6a=-\sqrt5/3$, you have $$ \cos 6a=\sqrt{1-(-\sqrt5/3)^2}=\sqrt{1-5/9}=\frac23 $$ (we choose the positive root because we are told that $\cos 6a>0$. Now from $\cos 2x=1-2\sin^2x$ we get $\sin x=\pm\sqrt{\frac{1-\cos 2x}2}$. To decide on the sign, we note that as $6a$ is in the fourth quadrant, its half should be in the second. Then $$ \sin 3a=\sqrt{\frac{1-\cos6a}2}=\sqrt{\frac{1-2/3}2}=\frac1{\sqrt6}. $$ And $$ \tan 12a=\frac{\sin 12a}{\cos12a}=\frac{2\sin6a\cos6a}{1-\sin^26a}=2\,\frac{-\sqrt5/3\times2/3}{1-2\times5/9}=4\sqrt5. $$ You should be able to do the other one.