Let \(L_1\), \(L_2\) be any two context-free languages and \(R\) be any…

Answer: I and III only.

• I. L1 ∪ L2 is context-free: True. Context-free languages are closed under union (you can combine grammars or use nondeterministic pushdown automata to accept the union).

• II. Complement of L1 is context-free: False in general. Context-free languages are not closed under complementation. For example, L = {a^n b^n | n ≥ 0} is context-free but its complement over {a,b}* is not context-free.

• III. L1 − R is context-free: True. L1 − R = L1 ∩ (complement of R). Regular languages are closed under complement, so complement of R is regular; intersection of a context-free language with a regular language is context-free.

• IV. L1 ∩ L2 is context-free: False in general. The class of context-free languages is not closed under intersection; for instance, intersecting {a^n b^n c^m} and {a^m b^n c^n} (both context-free) yields {a^n b^n c^n}, which is not context-free.

Therefore the only statements that are always true for arbitrary context-free L1, L2 and regular R are I and III.