Hardy Weinberg problem help!
Your answer to a) is correct (standard Hardy-Weinberg equation).

For b) you have to consider that you are calculating proportions of the population, which should sum to 1. Since 30% of the AA individuals and all SS individuals die, this is not the case after selection, which is why you should divide the frequencies by the total proportion that survive (also for the AS genotype), such as:

$$
\begin{array}{c|cccc}
\text{genotype} &amp; \text{freq before} &amp; w_{survival} &amp; \text{freq after selection} &amp; \text{corrected freq}\\
&amp;\text{selection}&amp;&amp;&amp;\text{after selection}\\
\hline
AA &amp; 0.64 &amp; 0.7 &amp; f(AA)\cdot w_{AA}=0.45 &amp; 0.58\\
AS &amp; 0.32 &amp; 1 &amp; f(AS)\cdot w_{AS}=0.32 &amp; 0.42 \\
SS &amp; 0.04 &amp; 0 &amp; f(SS)\cdot w_{SS}=0.0 &amp; 0.0\\
\text{sum} &amp; 1 &amp; &amp; 0.768 &amp; 1
\end{array}
$$

here $w$ is the fitness of genotypes. And for your comment on why you didn't have to divide by the sum of freqencies in a): you could do this, but since the sum of genotype frequencies is 1 it doesn't make a difference.

For c) you use the standard formula for calculating allele frequencies from genotypes, so that $f(A)=f(AA)+f(AS)/2= 0.79$ and $f(S)=f(SS)+f(AS)/2= 0.21$.