Hardy-Weinberg equilibrium describes the allele frequencies in a non-evolving population. It uses this equation:
p^2 + 2pq + q^2 = 1
p^2 is the frequency of homozygous dominant individuals. (AA)
2pq is the frequency of heterozygous individuals. (Aa or aA)
q^2 is the frequency of homozygous recessive individuals. (aa)
You got the right answer for 1, since you know that q^2=0.30
For 2, you have to set up a ratio. The numerator is the frequency of heterozygotes, or 2pq. To find the denominator, you have to limit the number of individuals you're talking about to only those who can recognize the bitter taste, i.e., those who are either AA or Aa. This is equal to p^2+2pq.
Your answer, the probability of carriers among people who can recognize the bitter taste, will be:
That's equal to 0.495/0.70, or .7071.
I hope this helps!