Let Σ = {𝑎, 𝑏, 𝑐}. For 𝑥 ∈ Σ∗ , and 𝛼 ∈ Σ, let #𝛼(𝑥) denote the number…

Answer: The regular languages among the choices are {a^m b^n | m, n ≥ 0} and { w ∈ {a,b}* | #a(w) ≡ 2 (mod 7) and #b(w) ≡ 3 (mod 9) }.

• {a^m b^n | m, n ≥ 0} is regular because it equals the regular expression a*b*. A DFA can be built with a start/accepting state that loops on a, transitions on the first b to a second accepting state that loops on b; any string of a's followed by b's is accepted.

• The set of strings where #a ≡ 2 (mod 7) and #b ≡ 3 (mod 9) is regular because only residues modulo fixed integers are constrained. A finite automaton can track (#a mod 7, #b mod 9) with at most 7×9 = 63 states and accept precisely the states with residues (2,3).

• Intersecting {a,b}* with {a^m b^n c^{m−n} | m ≥ n ≥ 0} removes any strings containing c, forcing m−n = 0 and yielding {a^m b^m | m ≥ 0}. This language is a classic non-regular language (can be shown by the pumping lemma), so the intersection is not regular.

• The language requiring #a ≡ 2 (mod 7) together with #a = #b remains non-regular because equality of the number of a's and b's is a non-regular property. Adding a modular constraint on #a does not remove the need for unbounded counting, so the language is non-regular (again, the pumping lemma can be used to prove this).