$\mathbb{Z_8}^*$ denotes the multiplicative group of $\mathbb{Z_8}$ as Mark Bennet has said.
You can show that $x \in \mathbb{Z_n}$ has a multiplicative inverse if and only if $(x,n)=1$. The proof is based on a special case of Bezout's theorem that states $(x,n)=1$ if and only if $\exists a,b \in \mathbb{Z}: ax+bn = 1$.
If $(x,n)=1$, then $\exists a,b: ax+bn=1$. This implies that $bn = 1-ax$ or $n \mid 1-ax$ which is the same as $ax \equiv 1 \pmod{n}$.
On the other hand, if there exists $a \in \mathbb{Z_n}$ such that $ax \equiv 1 \pmod{n}$ then $\exists b \in \mathbb{Z}: bn = 1 - ax$. Which gives you the converse.
So, the necessary and sufficient condition for an element in $\mathbb{Z_n}$ to be invertible is that it is relatively prime to $n$.
$(\Bbb Z/n\Bbb Z)^\times$ often means the group of units. It consists of all the elements in $\Bbb Z/n \Bbb Z$ that have an inverse. These elements form a group with multiplication.
Example:
$\Bbb Z/4\Bbb Z=\{0,1,2,3\}$ form a group with respect to addition $\langle\Bbb Z/4\Bbb Z, +\rangle$
To form a group with multiplication, with the same set, we need to throw out some elements. $2\in\Bbb Z/4\Bbb Z$ is bad, because it has no inverse, also $0$ has no inverse. We're left with $\{1,3\}$, so $(\Bbb Z/4\Bbb Z)^\times=\langle\{1,3\},\times\rangle$.
What does $\Bbb Z/4\Bbb Z$ mean?
From primary school some child might have written something like
$$8+7=11$$
A teacher would come to the child and say that he is wrong, and that the answer is $15$. If the child asks why, then the teacher might answer as follows:
To figure out the sum of two numbers we use a number line.
The expression $8+7$ means that we stand at the $8$-mark, and jump $7$ times to the right. The answer is where we land. Let's do this.
So the answer is indeed $15$. Now, what would happen to the numberline if $8+7$ were $11$? Addition with other numbers would also change:
$$\begin{align}8+7=11\\8+6=10\\8+5=9\\8+4=8\end{align}$$ This would all have to be true if $8+7=11$, in particularis the last line strange, $8+4=8$. This would mean that adding $4$ changes nothing, i.e. $$\boxed{4=0}$$
On the number line, this would mean that "jumping" $4$ times does not takes us to a new number. What would such a number line look like?
Let's see how many different number we operate with:
$$\begin{align}0=4=8=12=\ldots\\1=5=9=13=\ldots\\2=6=10=14=\ldots\\3=7=11=15=\ldots\end{align}$$
We are in fact only operating with $4$ numbers, with the rule that adding $4$ changes nothing. One could draw a such a number line like this:
But it would be much simpler to draw it like this:
This is "system", or numberline, to do arithmetic in is called $\Bbb Z/4\Bbb Z$, where the $4$ denotes that adding $4$ does not change a thing (it's an additive identity).
Best Answer
For $p$ prime, $\mathbb Z_p^*$ is used to denote the multiplicative group of nonzero elements of $\mathbb Z_p$.
On the other hand, $\mathbb Z_n^×$ denotes the group of units modulo $n$, which has order $\varphi (n) $, where $\varphi$ is Euler's totient function.
For prime $p$, we have $\varphi (p)=p-1$, and indeed $\mathbb Z_p^*=\mathbb Z_p^×$.