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

(i)
Since we have a geometric progression, $p = 128(\frac{2}{3})^{n-1}$

$\mathrm{ln}p = \mathrm{ln}[128(\frac{2}{3})^{n-1}]$

$\mathrm{ln}p = \mathrm{ln}128 + (n-1)\mathrm{ln}\frac{2}{3}$

$\mathrm{ln}p = \mathrm{ln}(2^{7}) + (n-1)(\mathrm{ln}2 - \mathrm{ln}3)$

$\mathrm{ln}p = 7\mathrm{ln}2 + n\mathrm{ln}2 - n\mathrm{ln}3 - \mathrm{ln}2 + \mathrm{ln}3$

$\mathrm{ln}p = (6+n)\mathrm{ln}2 + (1- n)\mathrm{ln}3$

$\therefore, A=1, B=6, C=-1, D=1$

(ii)
$S_{\infty} = \frac{128}{1-2/3}=384$.
Thus, the length cannot be greater than 384cm.

(iii)
$S_{n}~>~380$

$\frac{128[1-(2/3)^{n}]}{1-2/3}~>~380$

$384[1-(2/3)^{n}]~>~380$

$(\frac{2}{3})^{n}~\textless~\frac{1}{96}$

$n \mathrm{ln}\frac{2}{3}~\textless~\mathrm{ln}\frac{1}{96}$

$n~>~\frac{\mathrm{ln}\frac{1}{96}}{\mathrm{ln}\frac{2}{3}}=11.26$

Therefore, 12 pieces must be cut off.