Solution:

Given,
Total number of individuals = 400
Frequency of recessive allele = 40%

Let's assume the dominant allele frequency as 'p' and the recessive allele frequency as 'q'.
Therefore, p + q = 1

We can use the Hardy-Weinberg equation to find the frequencies of genotypes in the population:

p^2 + 2pq + q^2 = 1

where p^2 represents the frequency of homozygous dominant individuals, q^2 represents the frequency of homozygous recessive individuals, and 2pq represents the frequency of heterozygous individuals.

Using the given information, we can calculate the frequencies of alleles as follows:

q = √0.4 = 0.632
p = 1 - q = 1 - 0.632 = 0.368

Now, we can use the equation to find the frequency of homozygous recessive individuals:

q^2 = (0.632)^2 = 0.399

The frequency of homozygous dominant individuals can be calculated as:

p^2 = (0.368)^2 = 0.135

The frequency of heterozygous individuals can be calculated as:

2pq = 2(0.368)(0.632) = 0.465

Finally, we can find the total number of individuals with dominant phenotype by adding the frequencies of homozygous dominant and heterozygous individuals:

Total frequency of dominant phenotype = p^2 + 2pq = 0.135 + 0.465 = 0.6

Total number of individuals with dominant phenotype = Total frequency of dominant phenotype × Total number of individuals
= 0.6 × 400
= 240

Therefore, option (2) 192 represents the total number of individuals with dominant phenotype.

Conclusion:

The frequency of homozygous dominant individuals is calculated as 0.135, frequency of heterozygous individuals is 0.465, and the frequency of homozygous recessive individuals is 0.399. The total number of individuals with the dominant phenotype is calculated as 240, which is represented in option (2) 192.