Find the order of the cyclic subgroup of the given group generated by the indicated element $\mathbf{29.}$ The subgroup of $U_6$ generated by $\cos\frac{2\pi}3+i\sin\frac{2\pi}3.$
$\mathbf{30.}$ The subgroup of $U_5$ generated by $\cos\frac{4\pi}5+i\sin\frac{4\pi}5.$
$\mathbf{31.}$ The subgroup of $U_8$ generated by $\cos\frac{3\pi}2+i\sin\frac{3\pi}2.$
where $U_n = \{z \in \mathbb{C} : z^n = 1 \}$ (nth roots of unity)
For example in (29), I have to multiply $e^{\frac{2\pi i}{3}}$ three times to get back 1. So the order is 1. However I don't see how the number "6" under the U plays a role here.
Also for instance, for the subgroup of $U_8$ generated by $e^{\frac{5\pi i}{4}}$, the order is 8 because $(e^{\frac{5\pi i}{4}})^8 = 1$. The answer given was 


The 1st number which makes $ e^{\frac{5\pi i}{4}} = 1$ is 8 because the number must be either 2,4, or 8. 2 and 4 are not working so $|\langle e^{\frac{5\pi i}{4}}\rangle| = 8$


I don't understand how 2 and 4 come up here or even related here? 
Also, sorry for the inconvenience I caused
 A: It seems you are confusing some things from group theory. Most of your questions are answered from Lagrange's theorem. 
The order of a group is the number of elements in a group. The order of an element $a$ is the smallest positive integer $x$ for which $a^x = 1$. 
Lagrange's theorem tells us that the size of a subgroup $H$ of $G$ must divide the size of $G$. That is if $H\le G$ then we must have $|H|$ divides $|G|$. This in particular implies that the order of an element must also divide the order of the group because the set $\{a,\ a^2,\ \cdots,\ a^x\}$ forms a subgroup called the cyclic subgroup of $a$ denoted $\langle a\rangle$. This is what we mean by the subgroup generated by $a$.
For your first question, your group is $U_6$ the $6$th roots of unity. The order of this group is $6$ and so every subgroup must be of order $1,\ 2,\ 3,\ 6$. You can see that the order of your element is $3$ and so the subgroup generated by it must be of order $3$ as well. In particular the subgroup is in fact $U_3$. The number $6$ has significance here in that it limits the orders we must try.
For the second question, the order of the group is $5$, so what must the order of the element be?
And for the last, again you have to try the divisors of $8$ which are $1,\ 2,\ 4,\ 8$. Clearly the element is not $1$ itself, so it remains to try $2,\ 4,\ 8$. This is what your solution suggests.
A: You are given $ U_n = \{z \in \mathbb{C} : z^n = 1 \} $, so for (29), you have
$$ z^6=1=e^0 \implies z^6 = e^{i2k\pi} \implies z=e^{i2k\pi/6}= e^{ik\pi/3} \,,k=0,1,2\,,$$
where the fact $e^{i2k\pi}=1\,\,, k\in \mathbb{Z}$ has been used.
