Congruence of integers and primitive roots Lemma: In $\mathbb{Z}$, if $l\ge 1$, $p$ any prime, and $x\equiv y\pmod{p^l}$ then $x^p\equiv y^p\pmod{p^{l+1}}$. 
The proof is by Binomial theorem. Assume the Lemma. 

Suppose $x^{p^l}\equiv 1\pmod{p^a}$ but $x^{p^{l-1}}\not\equiv 1\pmod{p^a}$. 
Then $x^{p^{l+1}}\equiv 1\pmod{p^{a+1}}$ by Lemma. 
Q.1 Is it necessary that $x^{p^{l}}\not\equiv 1\pmod{p^{a+1}}$?

On the other hand, suppose that $x^{p^{l+1}}\equiv 1\pmod{p^{a+1}}$. 
Q.2  Is it necessary that $x^{p^{l}}\equiv 1\pmod{p^{a}}$?

I was thinking of these problems group-theoretically as problem regarding  primitive roots of unity, but I was unable to solve it group theoretically and even by number theory methods. Can one help me? 
I will be happy even if I get a small hint for problems. 
Note: If the problem is clear and if title is misleading, one may edit appropriate title.

If $p>2$ and $a>0$ then $x\equiv 1\pmod{p^a}\iff x^p\equiv 1\pmod{p^{a+1}}$ (thus, for $p>2$, the answers to your questions are both "yes"). Indeed, if $x$ satisfies either condition, then $[x^p\equiv]x\equiv 1\pmod p$, so we can consider $x=1+p^n y$, where $\color{blue}{n>0}$ and $p\not\mid y$. But then
$$x^p=1+p^{n+1}y+p^{2n+1}\underbrace{\left(p^{n(p-2)-1}y^p+\sum_{k=2}^{p-1}\frac{1}{p}\binom{p}{k}p^{n(k-2)}y^k\right)}_{\text{integer}},$$
i.e. $x^p=1+p^{n+1}z$ with $\color{blue}{p\not\mid z}[=y+p^n\cdot\ldots]$; the claim follows easily.
(For $p=2$ there must be $a>1$. "Analyse it".)
