Explanation

(a)
Since, $- 3 < x + \frac{2}{x} < 3$
$\Rightarrow - 3 < \frac{\left( x^{2} + 2 \right)x}{x^{2}} < 3$
$\Rightarrow - 3x^{2} < \left( x^{2} + 2 \right)x < 3x^{2}$ $(x \neq 0)$
$\Rightarrow x\left( x^{2} + 3x + 2 \right) > 0$
And $x\left( x^{2} - 3x + 2 \right) < 0\ \ \ \ \ \ (x \neq 0)$
$\Rightarrow x(x + 1)(x + 2) > 0$
And $x(x - 1)(x - 2) < 0$
$\Rightarrow x \in ( - 2, - 1) \cup (0,\infty)$ …(i)
And $x \in ( - \infty,0) \cup (1,\ 2)\ \ \$…(ii)
From relations (i) and (ii), we get
$x \in ( - 2, - 1) \cup (1,\ 2)$