Yes, it makes sense. The order of an element $g$ in some group is the least positive integer $n$ such that $g^n = 1$ (the identity of the group), if any such $n$ exists. If there is no such $n$, then the order of $g$ is defined to be $\infty$.
As noted in the comment by @Travis, you can take a small permutation group to get an example. For instance, the permutation $(1,2,3,4)$ in the symmetric group $S_4$ of degree $4$ (all permutations of the set $\{1,2,3,4\}$) has order $4$. This is because
$$(1,2,3,4)^1 = (1,2,3,4)\neq 1,$$
$$(1,2,3,4)^2 = (1,3)(2,4)\neq 1,$$
$$(1,2,3,4)^3 = (1,4,3,2)\neq 1$$
and
$$(1,2,3,4)^4 = 1,$$
so $4$ is the smallest power of $(1,2,3,4)$ that yields the identity.
For the additive group $\mathbb{Z}$ of integers, every non-zero element has infinite order. (Of course, here, we use additive notation, so to calculate the order of $g\in\mathbb{Z}$, we are looking for the least positive integer $n$ such that $ng = 0$, if any. But, unless $g = 0$, there is no such $n$, so the order of $g$ is $\infty$.)
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.
Best Answer
1) Z24 includes 0 as an element. 2) Your definiton of order is correct. 3) If an element has order 5, then x+x+x+x+x=0 mod 24, and this should be the smallest n auch that when x is added n times, it is 0. The equation implies x is 0 (e). But the order of e is 1. So there is no element of order 5. Also, you are working with the additive group mod 24 - the set is not a group under multiplication, so the identity is 0, and addition is "multiplication".