(a) ${\log _{0.2}}{{x + 2} \over x} \le 1$…..$(i)$

For log to be defined, ${{x + 2} \over x} > 0$ $ \Rightarrow $$x > 0$ or $x < – 2$

Now from $(i),$ ${\log _{0.2}}{{x + 2} \over x} \le {\log _{0.2}}0.2$

$ \Rightarrow $${{x + 2} \over x} \ge 0.2$ …..$(ii)$

Case $(i)$ $x > 0$

From $(ii),$ $x + 2 \ge 0.2x$

$ \Rightarrow $ $0.8x \ge – 2$

$ \Rightarrow $$x \ge – {5 \over 2}$.

Case $(ii)$ $x < – 2$

From $(ii),$ $x + 2 \le 0.2x$$ \Rightarrow $$0.8x \le – 2$$ \Rightarrow $$x \le – {5 \over 2}$

$ \Rightarrow $$x \in (0,\,\infty )\, \cup \,\left( { – \infty ,\, – {5 \over 2}} \right]$; 

$\therefore x \in \left( { – \infty ,\, – {5 \over 2}} \right] \cup (0,\,\infty )$.