a
(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 \((ii),\) \(x + 2 \le 0.2x\)\( \Rightarrow \)\(0.8x \le - 2\)\( \Rightarrow \)\(x \le - {5 \over 2}\)

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

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