From that we have:

 1 ∈ A,   1 ∉ B
 3 ∈ A,   3 ∉ B
 7 ∈ A,   7 ∉ B
11 ∈ A,  11 ∉ B 

B - A= {2, 6, 8}, 

From that we have:

 2 ∈ B,   2 ∉ A
 6 ∈ B,   6 ∉ A
 8 ∈ B,   8 ∉ A

A ∩ B = {4, 9}

From that we have:

 4 ∈ A,   4 ∈ B,   9 ∈ A,   9 ∈ B 

So we end up with

A = {1,3,7,11,4,9}, B = {2,6,8,4,9}

----------------------------------------

From that we have:

 1 ∈ C,   1 ∉ D
 2 ∈ C,   2 ∉ D
 4 ∈ C,   4 ∉ D

D − C = {7, 8}, 

From that we have:

 7 ∈ D,   7 ∉ C
 8 ∈ D,   8 ∉ C

C ∪ D = {1, 2, 4, 5, 7, 8, 9}

Since 5 and 9 are not in either C - D or in D - C,
they must be elements of both sets, since they were
removed in both cases of set subtraction.  Therefore

C = {1,2,4,5,9},  D = {7,8,5,9}

Edwin