Solution

The correct answer is Regular, Not regular

Key Points

       1. \( A = \{a, b\}\)
       2. \(L = A^*\)(the Kleene closure of  A , which includes all possible strings over  A , including the empty string)
       3. \( X = \{a^nb^n \mid n > 0\}\) (the set of strings with equal numbers of 'a's and 'b's, with at least one 'a' and one 'b') 

Now, let's consider \(L \cup X\), the union of languages L  and X :

        \(L \cup X = A^*\)

This language consists of strings with equal numbers of 'a's and 'b's, including the empty string. It is subset of A*. So, this language is regular, and its regular expression is \((a,b)^*\)

Now, let's consider X:

        \(X = \{a^nb^n \mid n > 0\}\)

This language consists of strings with equal numbers of 'a's and 'b's, with at least one 'a' and one 'b'. This language is not regular, as it cannot be recognized by a regular grammar. So, the correct answer is: 3) Regular, Not regular