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

(i)
$\frac{dz}{dx} = 3 - 2z$

$\frac{1}{3-2z} \frac{dz}{dx} = 1$

$\int \frac{1}{3-2z} dz = \int 1 dx$

$-\frac{1}{2} \mathrm{ln} |3-2z| = x + C$

$\mathrm{ln} |3-2z| = -2x + C$

$3-2z = e^{-2x +C}$

$2z = 3 - \frac{e^{-2x+C}}{2}$

$z = \frac{3}{2} - Ae^{-2x}$ where $A=\frac{e^{C}}{2}$

(ii)
$\frac{dy}{dx} = z$

$\frac{dy}{dx} = \frac{3}{2} - Ae^{-2x}$

$y = \frac{3}{2}x + \frac{Ae^{-2x}}{2} + D$

(iii)
$\frac{d^{2}y}{dx^{2}} = 2Ae^{2x}$

$\frac{d^{2}y}{dx^{2}}= 2(\frac{3}{2} - \frac{dy}{dx})$

$\frac{d^{2}y}{dx^{2}}= 3 - 2\frac{dy}{dx}$

$\therefore, a=-2, b = 3$

(iv)
Consider $A = 0$, then $y = \frac{3}{2} + D$

Let $D > 0$, so $y\frac{3}{2} + 1$ and $D=0$, then $y = \frac{3}{2}x$

$\therefore, \mathrm{they~are} y=\frac{3}{2}x ~\mathrm{and}~ y= \frac{3}{2} + 1$