Could somebody please verify or help me along with my answers? Thank you.
My updated work(maybe someone can help me a little more now): 
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Re: #6, what is $\cos(\pi/2)$? Note that for ANY $\theta$, we have: $|e^{i\theta}| = 1$, but $|1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$. – David Wheeler Mar 04 '15 at 01:14
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The first one is incorrect:
$$ e^{i\frac{\pi}{2}} = \cos(\pi/2) + i \sin(\pi/2) \neq 1 + i $$
The second one looks good.
The third and fourth ones you didn't attempt.
Hints: this is how you set these up. If $w = z_1^{z_2}$ where $z_1$ and $z_2$ are complex numbers, then you write $$\log w = z_2 \log z_1$$ so $$w = e^{\log w} = e^{z_2 \log z_1}.$$ (If you just want to take away the shortcut $z_1^{z_2} = e^{z_2 \log z_1}$ I guess that's fine, but I'd be remiss if I didn't at least wave my hands at the "why" of the matter).
For the last one, if $z = re^{i\theta}$, then $$\log z = \log{re^{i\theta}} = \log r + \log{e^{i\theta}} = \dots$$
BaronVT
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1sorry about that, I realized what was wrong with what I'm saying all I meant to say is if the angle is pi\2 then the x+yi form should consist only of yi because z should be 0. – Ali.B Mar 04 '15 at 01:30
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Okay, thanks for the help. Could you elaborate a little more on the third and fourth ones? I genuinely am struggling with this, not trying to be needy and get an answer out of you without trying. – Jeff Mar 04 '15 at 01:34
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Ok, @Ali.B makes sense. Jeff, follow Ali.B's suggestion to find $i^i$. – BaronVT Mar 04 '15 at 02:22
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@jeff actually, I suggest you do the final problem (#9) before #8, as it will be helpful to know what $\ln i$ is. (another hint: you need to find $w$ so that $e^w = i$. Can you think of an angle $\theta$ so that $e^{i\theta} = i$?) – BaronVT Mar 04 '15 at 02:25