(1) According to the selfing of purple flowers in C combination F 1, the sum of all the character coefficients in F2 is 16, which belongs to the special segregation ratio of 9: 3: 3: 1, so it can be inferred that the color of flowers is controlled by two pairs of genes (represented by A and A, B and B) and follows the genes. ② Safflower (aaB_) and White Flower (A_bb and aabb), for the convenience of calculation, according to the first case, F 1 purple flower in group A is selfed, and F _ 2 purple flower: safflower = 3: 1, indicating that the genotype of F 1 purple flower may be AABb, and further according to the parents in group A, all of them are

(2) In the same way, according to the first situation, the genotype of F 1 purple flower in group B may be AaBB, and the cross combination of parents is purple flower (AABB)× white flower (AABB). When the F 1 purple flower of group B and group C crosses, that is, AaBB×AaBb, then the phenotype and proportion of offspring are purple flower (3A_B_): white flower (66.

(3) F 1ABB in Group C produced F2 genotypes and proportions of 1ABB, 2aaabb, 2aaabb, 1ABB, 2aaabb, 1ABB, among which F2 selfed can produce safflower progeny, including 2aaabb, 4AaBb and/kl.

(4)① Select homozygous safflower to cross with group C white flowers, and then observe the phenotype of the offspring; ② If the offspring has both purple flowers and red flowers, it is a white-flowered hybrid plant.

So the answer is:

(1) Two (1) purple flowers × red flowers ((2), and the right one is 1).

(2) Purple flower: white flower = 3: 1 ((2 points), performance and proportion 1 minute)

(3) 1800(2 points)

(4)① Select homozygous safflower to hybridize with group C white flowers (2 points)? Observe the phenotype of offspring (1)

② There are both purple flowers and red flowers (2 points)