All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\frac{dy}{dx} = \frac{3x+4}{2\sqrt{x+2}}$

For the curve to have a turning point, $\frac{dy}{dx}=0$

$\Rightarrow 3x+4=0$

$x = -\frac{4}{3}$

(ii)
(a)
$y = \pm x \sqrt{x+2}$

$\frac{dy}{dx}= \pm \frac{3x+4}{2\sqrt{x+2}}$

When $x=0, \frac{dy}{dx}= \pm sqrt{2}$

(b)

(iii)