(a) To find the probability that a randomly chosen employee takes more than 28.6 minutes, we use the standard normal distribution. First, we calculate the z-score:

\(z = \frac{28.6 - 32.2}{9.6} = -0.375\)

We need \(P(Z > -0.375)\). Using the standard normal distribution table, \(\Phi(-0.375) = 0.3538\). Therefore, \(P(Z > -0.375) = 1 - 0.3538 = 0.6462\).

Thus, the probability is approximately 0.646.

(c) We need to find the probability that the time differs from the mean by less than 15.0 minutes, i.e., \(|X - 32.2| < 15\).

This is equivalent to \(17.2 < X < 47.2\). We calculate the z-scores:

\(z_1 = \frac{17.2 - 32.2}{9.6} = -1.5625\)

\(z_2 = \frac{47.2 - 32.2}{9.6} = 1.5625\)

We need \(P(-1.5625 < Z < 1.5625)\). Using the standard normal distribution table, \(\Phi(1.5625) = 0.9409\).

Thus, \(P(-1.5625 < Z < 1.5625) = 2 \times 0.9409 - 1 = 0.8818\).

Therefore, the probability is approximately 0.882.