Correct Option is (3) L_{1} is context free language, L_{2} is not context free language.

For $L_{1}$, consider the languages $L_{11} = \{ a^{n}b^{n} \}$ and $L_{12} = \{ a^{100} b^{100} \}$.

Here, $L_{11}$ is DCFL, and $L_{12}$ is regular. So their difference (i.e. $L_{11} \setminus L_{12}$) is context free.

Context-free languages are not closed under complement, intersection, or difference. This was proved by Scheinberg in 1960.^{[6]} However, if L is a context-free language and D is a regular language then both their intersection {$ L\cap D$} and their difference {$L\setminus D$} are context-free languages.