(i) To find the number of students expected to take longer than 20 minutes, we first standardize the variable:

\(P(X > 20) = P\left(Z > \frac{20 - 26.4}{3.7}\right) = P(Z > -1.730)\)

Using the standard normal distribution table, \(P(Z > -1.730) = 0.9582\).

For a sample of 350 students, the expected number is \(350 \times 0.9582 \approx 335 \text{ or } 336\).

(ii) A ‘very slow’ student is more than 1.645 standard deviations above the mean:

\(P(\text{very slow}) = P(Z > 1.645) = 0.05\).

We use the binomial distribution for 8 students with \(p = 0.05\):

\(P(0, 1, 2) = (0.95)^8 + 8 \times (0.05)^1 \times (0.95)^7 + \binom{8}{2} \times (0.05)^2 \times (0.95)^6\)

\(= 0.6634 + 0.2793 + 0.0515 = 0.994\).