in FA 2014 they say the frequency of an X linked recessive disease in males=q and in females =q2 ???????

That's right. Among females in XR-diseases calculations are the same as in AR-diseases. Because female may be a carrier or diseased (XHXh or XhXh) - like in AR: Aa or aa.

frequency of female-carrier (XHXh) = 2pq; of diseased female = q2.

But males-carriers in XR-diseases are diseased males at the same time (XhY, there's no another variant, he is ill and carrier in one). So the only unit that is used in calculations - q.

Try to repeat it in Kaplan lecture notes. This type of question I met in the Kaplan QBank.

May be the following discussion will be useful (I don't remember where it's from):

Let q be the frequency of the recessive affected allele and p be the frequency of the dominant unaffected allele. Thus, the female recessive homozygote genotype frequency is q2, so:

q2 = 9% = 0.09

q = 0.3

Since the allele is on the X chromosome, the frequency of affected males is equal to the allele frequency, so the frequency of affected males is 0.3 (or 30%). Note that this means that of the males, 30% of them will be affected.

Since I am assuming there are only two alleles, then the sum of the frequencies of the alleles is one:

p + q = 1

p = 1 - q

p = 1 - 0.3

p = 0.7

So, the frequency of unaffected males is 0.7 (or 70%).

The dominant homozygotes (and recessive homozygotes) and carriers will all be females (since they have two X chromosomes), and the frequency of the females that are carriers is given by 2pq:

2pq = 2 x 0.7 x 0.3 = 0.42

In order to determine the frequency of the population that are carriers, we would need to know the the frequencies of males and females in the population. Then, we could calculate the frequency of carriers in the whole population by multiplying the frequency of carriers by the percentage of the population that are females. Note that the above number (0.42) is simply the frequency of females that are carriers, not the frequency of carriers in the whole population.