(viii) A ∩ B' ∩ C'.

Here,

 S ={{(1,1), (1,2), (1,3), (1,4), (1,5), (1, 6), (2,1), (2,2), (2,3), (2,4), (2,5), (2, 6),(4,1), (4, 2), (4, 3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

A = {(2,1), (2,2), (2,3), (2,4), (2,5), (2, 6),(4,1), (4, 2), (4, 3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1, 6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5, 2), (5,3), (5,4), (5,5), (5,6)}

C = {(1, 1) (1,2), (1,3), (1,4), (2, 1), (2,2), (2, 3), (3, 1), (3,2), (4,1)}

(i) A' = S - A

where S is the sample space associated with the experiment of throwing two dice

i.e., S = A + B

Hence, A' = B

(ii) not B = S - B = A

(iii) A or B = (A ∪ B) = S

(iv) A and B = (A ∩ B) = φ

(v) A but not C = A - C

                      = {(2,4), (2,5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6,1), (6, 2) (6, 3), (6, 4), (6,5), (6,6)}

(vi) B or C = (B ∪ C)

               = {(1,1), (1,2),(1,3),(1,4),(1,5),(1,6), (2,1), (2,2), (2, 3), (3,1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4,1), (5,1), (5, 2), (5,3), (5, 4), (5, 5) (5, 6)}

(vii) B and C = (B ∩ C) = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3,2)}

(viii) B' = A

C’’= {(1,5), (2, 4), (2, 5), (2, 6), (3,3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4,4), (4,5), (4,6), (5,1), (5,6), (6,1), ....,(6,6)}

Now, A ∩ B' ∩ C = A ∩ A ∩ C = A ∩ C'

   = {(2,4), (2,6), (4,2), (4,3), (4,4), (4,5), (4, 6), (6,1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}