(i) To find the number of days a student uses the machine for less than 4 hours, we calculate \(P(X < 4)\). Standardize using \(Z = \frac{X - \mu}{\sigma}\), where \(\mu = 3.24\) and \(\sigma = 0.96\):

\(Z = \frac{4 - 3.24}{0.96} = 0.7917\)

Using the standard normal distribution table, \(P(Z < 0.7917) = 0.7858\).

Expected days: \(0.7858 \times 365 = 287\) days.

(ii) We need \(P(X > k) = 0.2\), which implies \(P(X < k) = 0.8\). Standardize:

\(P\left(Z < \frac{k - 3.24}{0.96}\right) = 0.8\)

From the standard normal table, \(Z = 0.842\).

\(\frac{k - 3.24}{0.96} = 0.842\)

\(k = 0.842 \times 0.96 + 3.24 = 4.05\)

(iii) We find \(P(-1.5 < Z < 1.5)\):

\(\Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1\)

\(\Phi(1.5) = 0.9332\), so \(2 \times 0.9332 - 1 = 0.866\).