Basically, you do not compute these lengths $2^n$ or $n^2$, but only the from-to computations that the automaton would be able to make on input with this particular lengths.
For the first two problems the automaton computes (in its internal states) the relation $E_\ell = \{ (p,q)\in Q\times Q \mid q\in\delta^*(p,y) \text{ for some $y$ with $|y|=\ell$} \}$ for some value of $\ell$ (which depends on $|x|$).
Then $E_{2\ell}=E^2_\ell$ which helps to solve the first problem. Here $E^2$ is the composition $E\circ E$ of the binary relation $E$ with itself: $E^2 = \{ (p,r) \mid \text{ for some } q\in Q \text{ both } (p,q)\in E \text{ and } (q,r)\in E\}$
I add some details. This is still rather formal, without much intuition. I hope it helps you into the right direction.
Assume we are given a FSA $\mathcal A= (Q,\Sigma,\delta,q_0,F) $ for language $A$. We will construct an automaton for your first problem.
The new states are of the form $(p,E)$ where $p\in Q$ and $E\subseteq Q\times Q$; i.e., the new state set equals $Q\times{\mathcal P}(Q\times Q)$.
Note that the number of possible relations $E$ is finite, hence adding such a relation to the states of the automaton will still lead to a finite state automaton.
The second component will always contain the relation $E_{2^n}$ where $n$ is the length of the input read.
The initial state equals $(q_0,E_1)$, where $E_1$ is the one-step relation on $Q$, just like defined above.
Now from every $(p,E)$ reading letter $a$ we move to state $(\delta(p,a),E^2)$. Thus, the first component copies the original move in $\mathcal A$, the second component steps from $E_{2^n}$ to $E_{2^n}\circ E_{2^n}=E_{2^{n+1}}$ following the recipe above.
When is a state $(p,E)$ final? When there is a pair $(p,q)\in E$ such that $q\in F$ is final in $\mathcal A$.
The $n^2$ problem is along similar lines, but actually slightly more involved (I think). It uses the fact that $(n+1)^2 = n^2 + 2n +1$. The automaton stores both $E_{n^2}$ and $E_{n}$ to easily update the square.
