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

(i)
From MF15,
$\mathrm{ln}(1+2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} + \ldots$

$\approx 2x - 2x^2 + \frac{8}{3}x^3$

(ii)
$ax(1+bx)^c = ax(1 + cbx + \frac{c(c)-1}{2!}(bx)^2 + \ldots)$

Comparing coefficients,

$a = 2$

$abc = -2 \Rightarrow bc = -1$

$2[\frac{c(c-1)}{2!}b^2] = \frac{8}{3}$

$\Rightarrow 2[\frac{c^2-c}{2c^2}] = \frac{8}{3}$

$c = -\frac{3}{5}$

$b = \frac{5}{3}$

Coefficient of $x^4 = a[\frac{c(c-1)(c-2)}{3!}(\frac{5}{3})^3] = -\frac{104}{27}$

The next parts, just need to be really careful. It is quite a lot of variables to juggle around here. For the last part, dont forget the $a$ and $b^2$