Welcome.
Cardinality means just the size of the set.
How many numbers are there in the set of natural numbers $\mathbb{N}$?
For $A$, just power both sides by $\frac{1}{4}$, is this a real number? and since $x$ is even powered, you have both positive and negative solutions.
For $B$, first you know that $n$ is a natural number, meaning $n = 1, 2, 3, 4, 5, \dots$. Just plug in $n$ into the equation of $x$, what $x$ do you get? From this $x$, which of them are also natural numbers?
Hope it helps,
Are you sure that this is supposed to be correct.
When $A$ and $B$ are sets with $A\cap B\neq\emptyset$, then there is $x\in A$ with $x\in B$.
Then $\{x\}\in\mathcal{P}(A)$. But $\{x\}\notin\mathcal{P}(A\setminus B)$. Since $\{x\}$ can not be a subset of $A\setminus B$.
Suppose otherwise, and $\{x\}\subseteq A\setminus B$. Then $x\in A\setminus B$, but $x\in A$ and $x\in B$. So $x\notin A\setminus B$. Which is a contradiction.
Edit: On your update.
When the statement is to show that $\mathcal{P}(A)=\mathcal{P}(A\setminus B)\Leftrightarrow~ A\cap B=\emptyset$, then we can prove this as follows:
$\Rightarrow$.
Let $\mathcal{P}(A)=\mathcal{P}(A\setminus B)$.
Suppose $A\cap B\neq\emptyset$. Then the proof given above can be applied here, to yield a contradiction.
You might want to work out the differences, or details yourself.
$\Leftarrow$:
Let $A\cap B=\emptyset$. Show $\mathcal{P}(A)=\mathcal{P}(A\setminus B)$.
This is trivial, as $A\cap B=\emptyset$ implies that $A\setminus B=A$.
Indeed:
$A\setminus B\subseteq A$ holds always.
For: $A\subseteq A\setminus B$ let $x\in A$. Since $A$ and $B$ are disjoint, we have that $x\notin B$. So $x\in A\setminus B$.
So we have shown that $A=A\setminus B$ (under the assumption that $A\cap B=\emptyset$.)
Best Answer
Backwards.
Since $A \subset B$, $A \cap B = A$
If $x \in A$,
So $x\in A \implies x\in B$ and hence, $A \subset B$.