(a) The probability that Tom removes a red disc on his first turn can be calculated by considering two scenarios: Sam removes a red disc first, then Tom removes a red disc, or Sam removes a white disc first, then Tom removes a red disc.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{2}{7}\)

Probability of Sam removing a white disc first: \(\frac{5}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{3}{7}\)

Total probability: \(\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7} = \frac{21}{56} = \frac{3}{8}\)

(b) Tom wins on his second turn if the sequence of draws is either RRWR, WRRR, or WRWR.

Probability of RRWR: \(\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRRR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRWR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{3}{5} = \frac{1}{14}\)

Total probability: \(\frac{1}{56} + \frac{1}{56} + \frac{1}{14} = \frac{3}{28}\)

(c) Given that Tom wins on his second turn, we need to find the probability that Sam removes a red disc on his first turn.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom winning on his second turn given Sam removes a red disc first: \(\frac{1}{56}\)

Total probability of Tom winning on his second turn: \(\frac{3}{28}\)

Conditional probability: \(\frac{\frac{1}{56}}{\frac{3}{28}} = \frac{1}{6}\)