abaR , a ∈ (0,1) *
We have to remember only one thing the strings generated should start and end with a same symbol.
The whole part of a and aR except the start and end symbol comes in b
(ii) In these expression the FA has to remember where a ends b starts and again where b ends and b starts so it not possible.
(iii) When b is fixed then abaR is not regular , where a ∈ (0,1) * . Because the whole string does not comes in b of a and aR.