By taking the sets of your own, verify that :

(i) n(A - B) = n(A ∪ B) - n(B)

(ii) n(A ∩ B) + n(A ∪ B) = n(A) + n(B)

Lets take the set A = {1, 2, 3, 4, 5, 6}

and set B = {2, 4, 6}

(i) n(A - B) = n(A ∪ B) - n(B)

A - B - contains all the elements in set A but not in B.

A - B = {1, 2, 3, 4, 5, 6} - {2, 4, 6}

A - B = {1, 3, 5}

n(A - B) = 3

A ∪ B - contains all the elements in set A and B.

A ∪ B = {1, 2, 3, 4, 5, 6} ∪ {2, 4, 6}

A ∪ B = {1, 2, 3, 4, 5, 6}

n(A ∪ B) = 6

n(B) = 3

Taking LHS : n(A - B)

n(A - B) = 3

Taking RHS : n(A ∪ B) - n(B)

n(A ∪ B) - n(B) = 6 - 3

n(A ∪ B) - n(B) = 3

∴ LHS = RHS

∴ n(A - B) = n(A ∪ B) - n(B)

(ii) n(A ∩ B) + n(A ∪ B) = n(A) + n(B)

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 4, 6}

A ∩ B = {2, 4, 6}

n(A ∩ B) = 3

A ∪ B = {1, 2, 3, 4, 5, 6}

n(A ∪ B) = 6

n(A) = 6

n(B) = 3

Taking LHS:

n(A ∩ B) + n(A ∪ B)

n(A ∩ B) + n(A ∪ B) = 3 + 6

n(A ∩ B) + n(A ∪ B) = 9

Taking RHS:

n(A) + n(B)

n(A) + n(B) = 6 + 3

n(A) + n(B) = 9

∴ LHS = RHS

∴ n(A ∩ B) + n(A ∪ B) = n(A) + n(B)