Interpretation of the notation $x = (x_1,x_2)\in \{0,1\}^2$? I have a few questions regarding the following notation:

$$
x = (x_1,x_2)\in \{0,1\}^2
$$

Question 1:
Is the following correct?
$\{0,1\}^2$ is the Cartesian product of the 2 sets $\{0,1\}$ and $\{0,1\}$, i.e.
\begin{align}
\{0,1\}^2 &= \{0,1\} \times \{0,1\} \\
&= \{(0,0),(0,1),(1,0),(1,1)\}
\end{align}
Question 2:
With the notation we mean $``$$(x_1,x_2)$ is an element of the set $\{0,1\}^2$$``$, so we can write:
$$
(x_1,x_2)\in \{(0,0),(0,1),(1,0),(1,1)\}
$$
So $(x_1,x_2)$ can take the values
\begin{align}
(x_1,x_2) &= (0,0)\\
(x_1,x_2) &= (0,1)\\
(x_1,x_2) &= (1,0)\\
(x_1,x_2) &= (1,1)
\end{align}
?
Question 3:
Does the notation mean that $(x_1,x_2)$ only can assign ONE value of $\{0,1\} \times \{0,1\}$? 
I.e. for $(x_1,x_2)$  we have 4 explicit cases:
\begin{align}
(x_1,x_2) &= (0,0) \\
\text{or} \quad
(x_1,x_2) &= (0,1)\\
\text{or} \quad 
(x_1,x_2) &= (1,0)\\
\text{or} \quad 
(x_1,x_2) &= (1,1)
\end{align}

Yes, yes and yes.
In general, the notation $A^n$, for a set $A$ and natural number $n$, means
$$
\underbrace{A \times \ldots \times A}_{n \text{ times}}.
$$
So $(x_1, \ldots, x_n) \in A^n$ means that each $x_i$, for $1 \leq i \leq n$, is an element of $A$. Thus this is a tuple of $n$ elements of $A$ (allowing duplicates).
