(i)

(ii)$\frac{-1-\sqrt{37}}{6}<x<\frac{1-\sqrt{5}}{2}$ or $1<x<\frac{1+\sqrt{5}}{2}$

(i)

(ii) Intersection between
$y=x+2+\frac{1}{x-1}$ and $y=2 x+2$
$2x+2 =x+2+\frac{1}{x-1} $
$x(x-1) =1 $
$x^2-x-1 =0 $
$x =\frac{1 \pm \sqrt{5}}{2}$

Intersection between
$y=x+2+\frac{1}{x-1}$ and $y=-2 x-2$
$-2 x-2 =x+2+\frac{1}{x-1} $
$-3 x-4 =\frac{1}{x-1} $
$(3 x+4)(1-x) =1 $
$-3 x^2-x+3 =0 $
$x =\frac{-1 \pm \sqrt{37}}{6}$
From the graph, the intersection occurs at $x<-1$. So, $x=\frac{-1-\sqrt{37}}{6}$

From the graphs in (i), the solution is $\frac{-1-\sqrt{37}}{6}<x<\frac{1-\sqrt{5}}{2}$ or $1<x<\frac{1+\sqrt{5}}{2}$