Solution

The correct answer is option 3.

Key Points

(a+b)* is a universal set grammer which can generate all strings over all possible alphabet "a"and"b".

(a* + b)^*= a* can generate epsilon, and reamaining part is like (a + b)*

(a^* + b)*=(a + b*)*=(a^*b^*)^*=(a^*+ b^*)^* =a^*(ba^*)^*=b^*(ab^*)^*=(a + b)* all grammers are equal.

∴ Hence the correct answer is L1 = L2.