**TRUE**?

$$\,\,\,\,\,\,\,{\rm I}.\,\,\,\,\,\,\,$$ $$\overline {{L_3}} \cup {L_4}$$ is recursively enumerable

$$\,\,\,\,\,{\rm I}{\rm I}.\,\,\,\,\,\,\,$$ $$\overline {{L_2}} \cup {L_3}$$ is recursive

$$\,\,\,{\rm I}{\rm I}{\rm I}.\,\,\,\,\,\,\,$$ $$L_1^ * \cap {L_2}$$ is context-free

$$\,\,\,{\rm I}V.\,\,\,\,\,\,\,$$ $${L_1} \cup \overline {{L_2}} $$ is context-free

$$\,\,\,$$ $${\rm I}.\,\,\,\,\,\,\,\,\,$$ 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?

For any two languages L_{1} and L_{2} such that L_{1} is context-free and L_{2} is recursively enumerable but not recursive, which of the following is/are necessarily true?

_{1}) is recursive

II. $${\overline L _2}$$ (complement of L

_{2}) is recursive

III. $${\overline L _1}$$ is context-free

IV. $${\overline L _1} \cup {L_2}$$ is recursively enumerable