(i) Let the probability of throwing an even number be \(p\). Then the probability of throwing an odd number is \(2p\). Since there are 3 even numbers (2, 4, 6) and 3 odd numbers (1, 3, 5), we have:

\(3p + 6p = 1\)

\(9p = 1\)

\(p = \frac{1}{9}\)

Thus, the probability of throwing a 3 is \(2p = \frac{2}{9}\).

(ii) The probability of throwing a 5 is \(\frac{2}{9}\) and the probability of throwing a 4 is \(\frac{1}{9}\). The probability of throwing two 5s and one 4 in three throws is:

\(\binom{3}{2} \left( \frac{2}{9} \right)^2 \left( \frac{1}{9} \right) = 3 \times \frac{4}{81} \times \frac{1}{9} = \frac{4}{243}\).

(iii) Let \(X\) be the number of times an even number is thrown. \(X\) follows a binomial distribution with \(n = 100\) and \(p = \frac{1}{3}\). The mean \(\mu = np = 33.3\) and the standard deviation \(\sigma = \sqrt{np(1-p)} = \sqrt{100 \times \frac{1}{3} \times \frac{2}{3}} = 22.2\).

Using a normal approximation, \(X \sim N(33.3, 22.2^2)\). We find \(P(X \leq 37)\) by standardizing:

\(P(X \leq 37) = P\left( Z \leq \frac{37.5 - 33.3}{22.2} \right) = P(Z \leq 0.8839) \approx 0.812\).