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

(i)
$(1+2i)^{3}$

$=(1+2i)^{2}(1+2i)$

$=[1+4i+(2i)^{2}](1+2i)$

$=(1+4i-4)(1+2i)$

$=(4i-3)(1+2i)$

$=4i-3+4i(2i)-3(2i)$

$=4i-3-8-6i$

$=-11-2i$

(ii)
Since $w=1+2i$ is a root,
$a(1+2i)^{3}+5(1+2i)^{2}+17(1+2i)+b=0$

$a(-11-2i)+5(-3+4i)+17(1+2i)+b=0$

$-11a-2ai-15+20i+17+34i+b=0$

$-11a+2+b-2ai+54i=0$

Comparing the real and imaginary coefficients,

$-2a+54 = 0 ~\Rightarrow ~ a=27$,

$-11(27)+2+b= 0 ~\Rightarrow ~ b=295$,

$\therefore a=27, b=295$

(iii)
Since all coefficient of the polynomial is real, by conjugate root theorem, if $1+2i$ is root, then $1-2i$ is also a root.
$(z-1-2i)(z-1+2i)=z^{2}-2z+5$

$27z^{3}+5z^{2}+17z+295 \equiv (z^{2}-2z+5)(cz+d)$,

by comparing coefficients, $c=27, d=59$

$\mathrm{Roots}=1+2i, 1-2i, -\frac{59}{27}$