Which of the following problems is/are decidable problem(s) (recursively…

Final answer: (a), (b) and (c).

Reasoning:

• For the emptiness of a context-free grammar (statement (a)): This is decidable. Compute the set of generating nonterminals (those that derive some terminal string) by iteratively adding nonterminals that produce terminals or previously found generating nonterminals. If the start symbol is not in that set, then L(G)=∅; otherwise L(G)≠∅.

• For the existence of two TMs whose union covers L(M) (statement (b)): This property is trivially true for every TM M: choose one of the two machines to be M itself and the other to accept the empty language (or any chosen language). Thus the existential statement holds for all inputs, so it is decidable (the decision is always "yes").

• For the tape-bound acceptance condition (statement (c)): Interpreted as the decision problem on pairs (M, ω): "Does M accept ω using at most 2|ω| tape cells?" this set is recursively enumerable. A semi-decision procedure is: simulate M on ω step-by-step while tracking the number of distinct tape cells visited; if M accepts and the number of used cells is ≤2|ω|, then accept. If M never accepts or accepts only after exceeding the bound, the simulation will not accept, which matches r.e. behavior.

Note: If statement (c) had instead been intended as a universal property over all inputs ω (i.e., "M accepts every ω using at most 2|ω| cells"), that variant is a different question and is not covered by the r.e. argument above and is typically undecidable. The usual interpretation for (c) in the context of language membership/decidability questions is the pairwise decision problem (M, ω).