Given the languages
L1={anb2m|n,m≥1}
L2={anb3n|n≥0}
L = L1 ∩ L2
I know that L1 is regular language and L2 can be represented by a PDA.
But I don't understand the answer which states that L is {a2nb6n|n≥1}. How was this solution computed?
A language is just a set (of valid strings), so what we have here is really just a simple problem in integer arithmetic. It's only necessary to remove the elegant costume of formal languages to arrive at the essence of the problem.
Both sets L1 and L2 are subsets of {acount(a)bcount(b)|j,k≥0}; that is, they consist of some number of as followed by some number of bs. The condition for L1 restricts count(b) to be a positive even number, since it must be 2m for some integer m≥1. The condition for L2 restricts count(b) to be exactly 3×count(a).
Now, the intersection of two sets defined with predicates is the set of elements such that both predicates are true. So count(b) must be both divisible by 2 and by 3, which means that it must be divisible by 6; in other words, it must be 6n for some positive integer n. And that means that count(a) must be 2n since there are exactly three times as many bs. That gives you the provided answer.
Like L2, L is not regular, but it can be implemented with a very similar PDA.