If $A = \{apples, oranges, bananas\}$ and $B= \{3,7,9\}$ then $A\cup B = \{apples, oranges, bananas,3,7,9\}$.
There is no restriction that if $A$ is a set of fruit that the only things you can do with it is in the realm of fruit nor that if $B$ is in the realm of numbers that the only things we can do with it are in the realms of numbers.
$A = \{4,5,6,7\}$ in the realm of single numbers.
And $B\times C =\{(6,4),(6,7),(6,9),(8,4),(8,7),(8,9)\}$ in the real of ordered pairs
And so $A \cup (B\times C) =\{4,5,6,7,(6,4),(6,7),(6,9),(8,4),(8,7),(8,9)\}$ which is not restricted to either the ream of number nor the realm of ordered pairs.
.....
As for relations. Note: one relation is subset of $A\times B$. A relation is not an element of $A\times B$.
Example if $A = \mathbb N$ and $B=\mathbb N$ and the relation is "the first number divides the second" then "the first number divides the second" $\ne (a,b)$ for any one pair $(a,b)$.
Instead "the first number divides the second" $= \{(1,2),(5,15), (7,35),,.......\}$ which is a whole subset of ordered pairs.
So if $C= \{1,5\}$ and $D=\{2,3,5\}$
Then what are the possible relations.
First is: $\emptyset$. That is no number is related to any other.
The second is $\{(1,2)\}$. That is $1$ is related to $2$ but nothing else are.
Another is $\{(1,2),(1,5), (5,3),(5,2)\}$. Which is a arbitrary $1$ is relateed to $2$ and $5$ and $5$ is related to $3$ and $2$. Why? Who cares?
The last is $\{(1,2),(1,3),(1,5),(5,2),(5,3),(5,5)\} = C\times D$. Everything is related to everything!
So how many total relations are there?
Well every subset of $C\times D$ can be a relation. so if $\mathscr P(C\times D)=\{$ the subsets of $C\times D\}$ then the number of relations is $|\mathscr P(C\times D)|$.
Which is a different concept then $|C\times D|$.
....
So you figure that if $|A| = k$ and $|B| =3$ then $|A\times B| = 3k$. That's fine.
But $4096 = |\mathscr P(A\times B)|$.
So you need to figure out if $|A\times B| = 3k$ then what does $|\mathscr P(A\times B)| = f(3k)$ is.
Can you?
Hint: If I eyeball factor $4096$ I get that $4|4096$ so $4096 = 4*1024$ and if I eyeball factor $1024$ I see it is divisible by $2$ so $4096 = 8*512$ and if I eyeball factor I get .... hey, wait a minute!
Best Answer
\begin{align}(x,y)\in (A\setminus B)\times(C\setminus D)&\iff x\in (A\setminus B)\wedge y\in(C\setminus D)\\&\iff x\in A\wedge x\notin B \wedge y\in C\wedge y\notin D\\&\iff(x,y)\in(A\times C)\wedge (x,y)\notin(B\times C)\wedge (x,y)\notin (A\times D)\\&\iff (x,y)\in (A\times C) \setminus [(A\times D)\cup (B\times C)] \end{align}