(i) (a) To get a final score of 12, the second throw must be 6 and the first throw must also be 6, since \(6 + 6 = 12\). The probability of this is \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).

(b) To get a final score of 5, the possible outcomes are (1,4), (2,3), (3,2), (4,1) where the second throw is not 5. The probability is \(\frac{4}{36} = \frac{1}{9}\).

(ii) Calculate probabilities:

\(P(A) = \frac{1}{6}\)

\(P(B) = \frac{5}{36}\) for pairs (1,5), (2,4), (3,3), (4,2), (5,1)

\(P(C) = 1 - P(O, O) = \frac{3}{4}\)

Check independence:

\(P(A \cap B) = \frac{1}{36}\) which is not equal to \(P(A) \times P(B)\)

\(P(A \cap C) = \frac{3}{36}\) which is not equal to \(P(A) \times P(C)\)

\(P(B \cap C) = \frac{2}{36}\) which is not equal to \(P(B) \times P(C)\)

None of the events are independent.