Hardy-Weinberg Color-blind
A: wild-type allele / a: color blind allele

Because color blindness is recessive and X-linked your assumption $p=F(a)=4\%$ is correct as men do only have one copy of the allele. Subsequently $F(A)=q=1-p=0.96$ is also correct. Therefore:

a) $F(Aa)=2pq=7.68\%$ is correct and b) is wrong, a is the color blind allele and $F(a)=0.04$ therefore it's $p^2=0.04^2=0.0016=0.16\%$.

92% color blind among females seems a bit high :).