$G$ is a group, if ∀a,b∈G, $a^2b=ba^2$, is $G$ Abelian? $G$ is a group, if ∀a,b∈G, $a^2b=ba^2$, how to make a counterexample show that $G$ is NOT Abelian?
What's the counterexample when $a^nb=ba^n$?

To get a counterexample, you need an element which is not the square of another element. A good example would be the quaternion group, where $i^2=j^2=k^2=-1$, and $(-1)^2=1$. $-1$ commutes with all the elements in the quaternion group.
Read the Wikipedia article here.
Since every element in the quaternion group raised to the $4$th power is the identity, it follows that for all $n|4$, your group does not need to be Abelian.
For $n\equiv2\mod4$, you can also use the quaternion group, as every element raised to the $n$th power would be either $-1$ or $1$, which commutes.
(This still leaves out the case when $n$ is odd.)
