Chance of child inheriting at least one copy of an allele - heterozygote mother, unknown father
It seems that what you are trying to do is calculate the probability of the child having at least one copy of the "risk allele" which I will denote as "A" (and it's counterpart "a").

First of all you know that the mother is Aa, and the child will get one allele at random. The probability that the mother passes on the A allele is then 0.5.

The tricky bit is the seemingly unknown genotype of the father. However, you do know genotype frequencies, which are the probability for the fathers three possible genotypes.


If the father is AA then it is certain the child will get the risk allele, and the probability of the father being AA is 0.0025. 
If the father is heterozygous then there is a probability of 0.5 he will pass on the A allele, and the probability of the father being Aa is 0.095. 
If the father is aa there is a probability if 0 that he will pass on the A allele. 


Combined these give the probability that the father carries at least one A and will pass it to his child.

$$(1 \times 0.0025) + (0.5 \times 0.095) + (0 \times 0.9025) = 0.05$$

The probability of the child getting at least one A is then the probability of getting it from the mother plus the probability of getting it from the father.

$$0.5 + 0.05 = 0.55$$