L = {a^n b^m | m = n&& m,n>=1}. Then we know it is not regular because we have to compare number of occurences of “a” with number of occurences of “b” . But if it is given that L = {a^n b^m | m = n && m,n>=1} is regular then m and n has to be finte. Otherwise we cant say L is regular.
Using the above argument we can say L is finite and L1,L2 are subset of L. Hence L1,L2 are also regular.