Answer on (a):Not in general. See the answers to this question.
Answer on (b):
If $ba^n=1=a^nb$ then $(ba^{n-1})a=1=a(a^{n-1}b)$ showing that $a$ has a left- _and_ a rightinverse. Consequently $a$ is invertible.
Answer on (a):Not in general. See the answers to this question.
Answer on (b):
If $ba^n=1=a^nb$ then $(ba^{n-1})a=1=a(a^{n-1}b)$ showing that $a$ has a left- _and_ a rightinverse. Consequently $a$ is invertible.