Let $${L_1} = \left\{ {{0^{n + m}}{1^n}{0^m}\left| {n,m \ge 0} \right.} \right\},$$
$$\,\,\,{L_2} = \left\{ {{0^{n + m}}{1^{n + m}}{0^m}\left| {n,m \ge 0} \right.} \right\},$$ and
$$\,\,\,\,{L_3} = \left\{ {{0^{n + m}}{1^{n + m}}{0^{n + m}}\left| {n,m \ge 0} \right.} \right\},$$ Which of these languages are NOT context free?

Consider the following statements about the context-free grammar
$$G = \left\{ {S \to SS,\,S \to ab,\,S \to ba,\,S \to \varepsilon } \right\}$$
$$1.$$ $$G$$ is ambiguous
$$2.$$ $$G$$ produces all strings with equal number of $$a's$$ and $$b's$$
$$3.$$ $$G$$ can be accepted by a deterministic $$PDA$$.

Which combination below expresses all the true statements about $$G?$$

For $$s \in {\left( {0 + 1} \right)^ * },$$ let $$d(s)$$ denote the decimal value of $$s(e. g.d(101)=5)$$
Let $$L = \left\{ {s \in {{\left( {0 + 1} \right)}^ * }\left| {\,d\left( s \right)\,} \right.} \right.$$ mod $$5=2$$ and $$d(s)$$ mod $$\left. {7 \ne 4} \right\}$$

Which of the following statement is true?

Let $${L_1}$$ be a regular language, $${L_2}$$ be a deterministic context-free language and $${L_3}$$ a recursively enumerable, but not recursive, language. Which one of the following statement is false?