L1={wxwR∣w,xϵ{a,b}∗and∣w∣,∣x∣>0}
w cannot be put as "ϵ " since ∣w∣>0. So we put w as its smallest string which is 'a' and 'b' and get the regular expression:
r=a(a+b)+a+b(a+b)+b.
Now putting w as any other string like say "ab" will not add any new string to the expression r. since any such string so generated will be either already generated by either a(a+b)+a or b(a+b)+b
So the given language = a(a+b)+a+b(a+b)+b which is clearly regular.
∴L1isregular
L2={anbm∣m≠n}. This language has infinite comparision between number of a's and b's. L2 is not regular language.
L3:{apbqcr∣P,q,r≥0} = a∗b∗c∗ is a regular language.