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

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

$z^{2}=1+4i-4=-3+4i$

$\frac{1}{z^{3}}=\frac{1}{(1+2i)(-3+4i)}$

$\frac{1}{z^{3}}=\frac{1}{-11-2i}\times \frac{-11+2i}{-11+2i}$

$\frac{1}{z^{3}}=\frac{-11+2i}{121+4}$

$\frac{1}{z^{3}}=\frac{-11}{125}+\frac{2i}{125}$

(ii)
Using (i)
$pz^{2}+\frac{q}{z^{2}}=p(-3+4i)+q(\frac{-11+2i}{125})$

Since it is real, $4p+\frac{2q}{125}=0$

We have $q=-250p$,

$pz^{2}+\frac{q}{z^{2}}=-3p-(\frac{11}{125})(-250p)=19p$