Regular

Question

how is{ w(w^R)* | w belongs to {0,1}* } regular?

Answer 1

For identifying language if you are not sure, go to basics and try brute force.
Is $\epsilon$ in $L$?
Is $0$ in $L$?
etc.
If we see $L$ is nothing but $\Sigma^*$, because $^*$ implies 0 or more occurrence. So, with 0 occurrence $L$ is containing all possible strings, so there is no need to evaluate for other repetitions. Now, try replacing $^*$ with $^+$ and see what language it is - if you have enough free time :)

Comments

@Arjun
With + it's not regular because we can't disappear second half(w^r) of the string, please confirm!

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yes, then CFL?

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no it's not CFL, we can't check all the powers of n where w(w^r)^n n>0 . it should be  CSL?

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it should be csl as we have to check for one or more occurrences of n