There are two ways of doing this. The first is essentially what you have done $\displaystyle sin(\pi/6)= \frac{1}{2}$ and $\displaystyle cos(\pi/6)= \frac{\sqrt{3}}{2}$ so that $\displaystyle z= \frac{3\sqrt{3}}{2}+ i\frac{3}{2}$. $\displaystyle \overline{z}= \frac{3\sqrt{3}}{2}- i\frac{3}{2}$. All that has to be multiplied. It is simpler if you realize that $\displaystyle (a+ bi)(a- bi)= a^2+ b^2$, a real number.
The other way is to use the exponential form. saying $\displaystyle z= 3(cos(\pi/6)+ i sin(\pi/6))$ is the same as saying $\displaystyle z= 3e^{i\pi/6}$. And then $\displaystyle \overline{z}= 3e^{-i\pi/6}$. And then the product is $\displaystyle (3)(3)(e^{i\pi/6- i\pi/6})= 9$
$\displaystyle \frac{8e^{\frac{\pi i}{2}}}{4e^{\frac{\pi i}{4}}}= \frac{8}{4}e^{\frac{\pi i}{2}- \frac{\pi i}{4}}= 2e^{\frac{\pi i}{4}}$
First if you do not know that 2/8= 1/4 you should give up all math!I got the answer 2e^(2pi/8)i
But the correct answer is 2+i sq.rt2
How do I get that answer?
And, of course, $\displaystyle 2e^{\pi i/4}= 2(cos(\pi/4)+ i sin(\pi/4))$. What is that equal to? (It is NOT "2+ i sq.rt 2"! I suspect you have misread that.)