For any two languages L₁ and L₂ such that L₁ is context-free and L₂ is recursively enumerable but not recursive, which of the following is/are necessarily true?

I. $${\overline L _1}$$ (complement of L₁) is recursive
II. $${\overline L _2}$$ (complement of L₂) is recursive
III. $${\overline L _1}$$ is context-free
IV. $${\overline L _1} \cup {L_2}$$ is recursively enumerable

Consider the following statements.

$$\,\,\,$$ $${\rm I}.\,\,\,\,\,\,\,\,\,$$ The complement of every Turing decidable language is Turing decidable
$$\,$$ $${\rm II}.\,\,\,\,\,\,\,\,\,$$ There exists some language which is in $$NP$$ but is not Turing decidable
$${\rm III}.\,\,\,\,\,\,\,\,\,$$ If $$L$$ is a language in $$NP,$$ $$L$$ is Turing decidable

Which of the above statements is/are true?

Let $$A\,\,{ \le _m}\,\,B$$ denotes that language $$A$$ is mapping reducible (also known as many-to-one reducible) to language $$B.$$ Which one of the following is FALSE?

Which of the following statements is/are FALSE?
$$1.$$ For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine
$$2.$$ Turing recognizable languages are closed under union and complementation
$$3.$$ Turing decidable languages are closed under intersection and complementation
$$4.$$ Turing recognizable languages are closed under union and intersection