Given the following two languages : \(L_{1} = \{uww^{R} ν | u, v, w \in (a,…

Conclusion: The first language is regular and the second language is not regular.

Reason L1 is regular: For any nonempty w, the string w w^R has the last character of w followed immediately by the same character (the first character of w^R). Thus w w^R always contains either "aa" or "bb". Therefore any string of the form u w w^R v contains "aa" or "bb" as a substring, so

L1 = (a+b)* (aa + bb) (a+b)*, which is a regular language.

Reason L2 is not regular: The constraint |u| ≥ |v| compares lengths of substrings before and after the central palindrome, which cannot be enforced by a finite automaton. A clear way to prove non-regularity is to intersect with a suitable regular language and apply the pumping lemma.

• Consider L2 ∩ a* bb a* = { a^p bb a^q | p ≥ q ≥ 1 }.

• Assume this intersection is regular. Let n be the pumping length. Take s = a^n bb a^n, which is in the intersection.

• Any decomposition s = xyz with |xy| ≤ n forces y to consist only of a's from the left block. Pumping down (i = 0) yields xz = a^{n-|y|} bb a^n, where the number of a's before bb is now less than the number after bb, so xz is not in the intersection. This contradicts the pumping lemma.

• Thus the intersection is not regular, so L2 cannot be regular.

Therefore the correct statement is: the first language is regular and the second language is not regular.