Finding the intersection of $A \times B$ and $B \times A$ $
\newcommand{\set}[1]{\left\{ #1 \right\}}
$Let $A = \set{x, 3}$ and $B = \set{y, 3, z}$. List all of the elements in the set
$(A × B) ∩ (B × A)$.

Is the Cartesian product of the two sets listed above a set of $2$-ordered tuples? And the intersection is just the elements that $A \times B$ has in common with $B \times A$? That is,
$$\begin{align*}
A \times B &= \{x,3\} \times  \{y,3,z\} = \{(x,y),(x,3),(x,z),(3,y),(3,3),(3,z)\} \\
B \times A &= \{y,3,z\} \times  \{x,3\} = \{(y,x),(y,3),(3,x),(3,3),(z,x),(z,3)\}
\end{align*}$$
giving $(A \times B)∩(B\times A) = \set{(3,3)}$
Is there some higher order of precedence? And I assume order matters, right? Any help would be greatly appreciated.
 A: Throughout, I assume $x,y,z$ are distinct items.

Is the cartesian product of the two sets listed above a set of $2$-ordered tuple?

Indeed.

And the intersection is just the elements that $A \times B$ has in common with $B \times A$?

Correct.

$$\begin{align*}
A \times B &= \{x,3\} \times  \{y,3,z\} = \{(x,y),(x,3),(x,z),(3,y),(3,3),(3,z)\} \\
B \times A &= \{y,3,z\} \times  \{x,3\} = \{(y,x),(y,3),(3,x),(3,3),(z,x),(z,3)\}
\end{align*}$$

These, too, are correct.

$$(A \times B) \cap (B \times A) = \{(3,3)\}$$

This is correct as well.

And I assume order matters, right?

In an ordered pair, order does matter. So $(x,y)$ and $(y,x)$ are different things, on the presumption $x \ne y$.
On the other hand, within a set, order does not matter. So $\{x,y\}$ and $\{y,x\}$ are the same sets.
As you have seen, too, the Cartesian product of sets is not necessarily commutative as well. So you might have that $A \times B$ and $B \times A$ give you different sets, as you have seen above.

Is there some higher order of precedence?

This, however, I'm not sure what you mean by, and so I can't answer it.
