(ii) For a normal approximation to be appropriate, both the expected number of successes and failures should be at least 5. Here, the expected number of successes is given by:

\(12 \times 0.25 = 3\)

Since 3 is less than 5, it is not appropriate to use a normal approximation.

(iii) Let the random variable \(X\) represent the number of days Martin is playing games when his friend phones in a 40-day period. \(X\) follows a binomial distribution with parameters \(n = 40\) and \(p = 0.25\).

The mean \(\mu\) and variance \(\sigma^2\) of \(X\) are:

\(\mu = 40 \times 0.25 = 10\)

\(\sigma^2 = 40 \times 0.25 \times 0.75 = 7.5\)

Using a normal approximation, we find \(P(X \geq 13)\) with continuity correction:

\(P(X \geq 13) = P\left( Z > \frac{12.5 - 10}{\sqrt{7.5}} \right)\)

\(= P(Z > 0.913)\)

Using the standard normal distribution table:

\(= 1 - \Phi(0.913)\)

\(= 1 - 0.8194\)

\(= 0.181\)