(i) To find the probability that Peter takes longer than 10.2 minutes, we standardize the variable:

\(P(x > 10.2) = P\left( z > \frac{10.2 - 9.5}{1.3} \right)\)

\(= P(z > 0.53846)\)

Using the standard normal distribution table, \(P(z > 0.53846) = 1 - 0.7046 = 0.295\).

(ii) To find the number of days Peter takes less than 8.8 minutes, we use symmetry:

\(P(x < 8.8) = 0.2954\)

Estimate the number of days: \(365 \times 0.2954 = 107.671\)

Rounding gives approximately 107 or 108 days.