Find the order of each element in $(\mathbb{Z}/18\mathbb{Z},+)$ and $((\mathbb{Z}/18\mathbb{Z})^{*},\cdot)$ Find the order of each element in $(\mathbb{Z}/18\mathbb{Z},+)$ and $((\mathbb{Z}/18\mathbb{Z})^{*},\cdot)$.
Using the definition of the order for a additive group: 

For $(G,+)$ a group and $x\in G$ define the order of $x$ to be
  the smallest positive integer $n$ such that $nx=0_G$, and denote this
  integer by $|x|$.

Then, there are elements $\bar{1}$, $\bar{2}$, ..., $\bar{17}$. $\bar{1}$ has order $18$, $\bar{2}$ has order $9$, $\bar{3}$ has order $6$, $\bar{4}$ has order $9$, $\bar{5}$ has order $10$, etc. Is this correct? And is it the same logic with $((\mathbb{Z}/18\mathbb{Z})^{*},\cdot)$ only using the definition:

For $G$ a group and $x\in G$ define the order of $x$ to be the
  smallest positive integer $n$ such that $x^n=1$, and denote this
  integer by $|x|$.  

I was given a hint to start by listing the elements of $\mathbb{Z}/18\mathbb{Z}$ that have a multiplicative inverse, but I don't know why.
 A: First, I hope you understand that the two definitions in your post are actually the same definition of the same concept, with the only difference being the notation chosen for the group operation — additive or multiplicative.
Second, as a minor omission, your list of element of $(\mathbb{Z}/18\mathbb{Z},+)$ is missing $\bar{0}$. Also, the order or $\bar{5}$ is $18$, not $10$, but it's probably a typo. Overall, you understand that one correctly.
Third, as for your last question, it's because of the very definition of what $((\mathbb{Z}/18\mathbb{Z})^{*},\cdot)$ is. That asterisk means the set of all invertible (with respect to multiplication) elements of the given ring. If we want to use multiplication as the group operation, then to actually have a group, each element has to have a multiplicative inverse — it's one of the group axioms. So elements that don't have inverses will have to be discarded.
A few examples. Obviously, $\bar{0}\notin(\mathbb{Z}/18\mathbb{Z})^{*}$. But $\bar{2}\notin(\mathbb{Z}/18\mathbb{Z})^{*}$ either. One way to see that is to notice that $\bar{2}\cdot\bar{9}=\bar{0}$, so $\bar{2}$ (as well as $\bar{9}$) is a zero-divisor, and therefore can't have an inverse. Or you can test all elements of $\mathbb{Z}/18\mathbb{Z}$ to see that none of them works as the inverse of $\bar{2}$.
So the hint simply tells you to actually right down all elements of $(\mathbb{Z}/18\mathbb{Z})^{*}$ to see what you're dealing with. You should see that discarding all zero-divisors modulo $18$ results in keeping only the elements that are relatively prime to $18$. And then by repetetive multiplication you can find the (multiplicative) order of each of them.
