Hardy Weinberg Equilibrium problem
This is basically the same solution as @AlanBoyd's answer, but since he asked me to, I will post my solution as well (also a reason to try out $\mathcal{MathJax}$ a bit more).

Assuming:

$f(A)=p=0.3, \\f(a)=q=0.7$ 

and the standard formulas:

$f(AA)=p^2\\
f(Aa)=2pq\\
f(aa)=q^2$

gives these genotypes in the parental generation (G1), before and after selection (using the fitness vector $w$):

$$
\begin{array}{c|cccc}
\text{genotype} &amp; \text{freq G1} &amp; w &amp; \text{freq breed.pop} &amp; \text{freq breed. rescaled}\\
\hline
AA &amp; 0.09 &amp; 1 &amp; f(AA)\cdot w_{AA}=0.09 &amp; 0.119\\
Aa &amp; 0.42 &amp; 1 &amp; f(Aa)\cdot w_{Aa}=0.42 &amp; 0.556 \\
aa &amp; 0.49 &amp; 0.5 &amp; f(aa)\cdot w_{aa}=0.245 &amp; 0.325\\
\text{sum} &amp; 1 &amp; &amp; 0.755 &amp; 1
\end{array}
$$

The genotype frequencies in the breeding population (rescaled so that they sum to one) are then used to calculate allele frequencies as:

$$f(A)=f(AA)+\frac{f(Aa)}{2} = 0.397\\
f(a)=f(aa)+\frac{f(Aa)}{2} = 0.603$$

These are then used to calculate the genotype frequencies in the next generation (G2), using the same HW-equations as in the beginning:

$$\begin{align}
f(AA)&amp;=0.397^2=0.158 \\
f(Aa)&amp;=2\cdot0.397\cdot0.603=0.479 \\
f(aa)&amp;=0.603^2=0.363 \\
\end{align}
$$

I do not understand where the given answer $f(Aa)=0.435$ is coming from.