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L1 = {0n+m1k+l | m = l, m, n, k, l >=1}
We can write it as
L1 = {0n0m1k1l | m = l, m, n, k, l >=1},Since m and l does not depend upon n and k, we can fix its value as 1, and use n and k to take remaining 0's and 1's, so our language become..
L1 = {0n01k1 |  n, k >=1}Now we just need to check for any number of 0 followed by a 0, then any number of 1 followed by a 1, which is regular.

L2 = {0n(11)m | m,n >=0}
Any number of zeros followed by even number of 1's, its regular.

According to me option D is correct, both L1 and L2 are regular.

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