Answer:
1. During meiosis, we have three chromosomes +, +, ey, when duplicating them in sister chromatids we will have the combination ++, ++, eyey, then multiple alignments are made to generate the metaphase, the pairs will be randomly joined and the odd will be left couple, two of the three arrangements will be added and will generate two sperm + ey and two +, on the other side 2 with ++ and 2 with ey will be produced, thus the genotypes will be the following:
1/3 +
1/3 hey
1/6 ++
1/6 hey
2.
to evaluate the probabilities of the genotype e and when crossing the spermatozoa already studied it will be the following
1/6 ++ hey
1/6 eyey
1/3 + eyey
1/3 + hey
With this result we can deduce that 1/6 will not have eyes and 5/6 are wild type, which supposes that there is no crossing and that wild flies have the genotype + eyey