(i) The probability of throwing a 1 is equal to the probability of throwing a 2, 3, 4, or 6. Since the probability of throwing a 5 is 0.75, the probability of throwing any other number is:

\(1 - 0.75 = 0.25\)

There are 5 numbers (1, 2, 3, 4, 6) with equal probability, so the probability of throwing a 1 is:

\(\frac{0.25}{5} = 0.05\)

The probability of throwing a 5 is 0.75, and the probability of throwing an even number (2, 4, 6) is:

\(\frac{3}{5} \times 0.25 = 0.15\)

The probability of the sequence 1, 5, even is:

\(0.05 \times 0.75 \times 0.15 = 0.00563\)

(iii) Let \(X\) be the number of times a 5 is thrown in 90 trials. \(X\) follows a binomial distribution \(B(90, 0.75)\).

Using a normal approximation, \(X \sim N(\mu, \sigma^2)\) where:

\(\mu = 90 \times 0.75 = 67.5\)

\(\sigma^2 = 90 \times 0.75 \times 0.25 = 16.875\)

We use a continuity correction for \(P(X > 60)\):

\(P(X > 60) = 1 - \Phi \left( \frac{60.5 - 67.5}{\sqrt{16.875}} \right) = \Phi(1.704)\)

\(\Phi(1.704) \approx 0.956\)