The first step is to write as much as genotype from the phenotype as possible.
Cross 1:
A/-; B/- X a/a; b/b = 1 A/-; B/-: 2 ?/a; ?/b: 1 a/a; b/b
Since there is a double recessive the blue parent should be A/a; B/b and the half purple should be A/a; b/b and a/a; B/b. respectively.
Cross 2:
?/?; ?/? X ?/?; ?/? = 3 A/-; B/-: 1 ?/?; ?/?
The two parents should be in the order A/a; b/b and a/a; B/b while the purple progeny should be the same. The blue progeny are A/a; B/b.
Cross 3:
A/-; B/- X A/-; B/- = 3 A/-; B/-: 1 ?/?; ?/?
It should be noted that from this cross we could say that one parent should be A/A or B/B and the other parent is B/b if the first is A/A or A/a if the first is B/B.
Cross 4:
A/-; B/- X ?/?; ?/? = 3 A/-; B/-: 4 ?/?; ?/?: 1 a/a; b/b
Either A/a; b/b or a/a; B/b could be the purple parent. Assume the purple parent is A/a; b/b and the blue parent must be A/a; B/b giving the following progeny.
3/4 A/- X 1/2 B/b = 3/8 A/-; B/b blue
1/2 b/b = 3/8 A/-; b/b purple
1/4 a/a X 1/2 B/b = 1/8 a/a; B/b purple
1/2 b/b = 1/8 a/a; b/b scarlet
Cross 5:
A/-; b/b X a/a; b/b = 1A/-; b/b:1 a/a; b/b
The 1:1 ratio denotes that it is a test cross for the gene A. The purple parent and the progeny are A/a; b/b. Also the purple parent and progeny could also be a/a; B/b.