$2A = \{ xy \mid x,y \in A \text{ and } x = y \} \rightarrow$ Not Regular (we can prove using Pumping lemma or Myhill-Nerode Theorem)
To prove that $2A$ is not regular, we use the Pumping Lemma proof by Contradiction here.
Assume $2A$ is regular.
Then according to the Pumping Lemma, there exists a pumping length $p \ge 1$ such that any $w \in 2A$ with $|w| \ge p$ can be written as $w = xyz$ satisfying:
1) $|xy| \le p$
2) $|y| > 0$
3) $xy^i z \in L$ for all $i \ge 0$
Let us assume $\Sigma = \{0,1\}$ and the string $w = \underbrace{1^p0}_{x} \underbrace{1^p0}_{y}$.
Its total length is $2p + 2 \ge p$.
Since $|xy| \le p$, the substring $y$ must lie entirely within the first block $1^p$.
Let $y = 1^k$ where $1 \le k \le p$.
Now, $ xy^i z = 1^{p + (i-1)k} 0 1^p 0 $
Taking $i = 2$, we get: \( xy^2 z = 1^{p+k} 0 1^p 0\)
Since the number of $1$'s in the first half is not equal to the number of $1$'s in the second half,
$xy^2 z \notin L$ for $1 \le k \le p$.
This contradicts the Pumping Lemma.
Therefore, $2A$ is not regular.
$A^2 = \{ xy \mid x,y \in A \}\rightarrow$ Regular (we can prove using FA)
As $A$ is regular we can have a finite automata for $A$, say that automata is $D$.
We first make 2 copies of this FA and name it $D_1$ and $D_2$.
From the final state(s) of $D_1$ we add $\epsilon$-transitions to the initial state of $D_2$.
We then make the final states of $D_1$ non-final.
The resulting NFA consisting of all the states and transitions of $D_1$ and $D_2$ alongwith the moidifications will accept the language $AA$ or $A^2$.
(This above statement can be verified formally.)
As a finite automata exists which accepts $A^2$, we can say it is a regular language.
