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(i)
We have that $R_{g} = (-\infty, \infty) \mathrm{and} D_{f} = \{ \mathbb{R} | x \ne 1 \}$,
Since $R_{g} \nsubseteq D_{f}, fg$ does not exist.

(ii)
$gf(x) = g( \frac{2+x}{1-x})$

$= 1 - 2(\frac{2+x}{1-x})$

$= \frac{1-x-4-2x}{1-x}$

$= \frac{3x+3}{x-1}$

Let $(gf)^{-1}(5) = a$

$gf[(gf)^{-1}(5)] = gf(a)$

$5 = \frac{3+3a}{a-1}$

$5a - 5 = 3 + 3a$

$a = 4$