(i) If the coin shows a head, the smallest sum of two dice throws is 2 (1+1). Therefore, X = 1 can only occur if the coin shows a tail and the first die shows 1. The probability of a tail is \(\frac{1}{2}\) and the probability of rolling a 1 is \(\frac{1}{4}\). Thus, \(P(X = 1) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\).

(ii) For X = 3, if the coin shows a head, possible outcomes are (1,2) and (2,1), each with probability \(\frac{1}{16}\). If the coin shows a tail, the first die must show 3, with probability \(\frac{1}{8}\). Therefore, \(P(X = 3) = \frac{2}{32} + \frac{1}{8} = \frac{3}{16}\).

(iii) Completing the table:

(iv) Events Q and R are exclusive because if a tail is thrown, X cannot be 7. Thus, \(P(Q \cap R) = 0\).