(a) We know that \(P(X > 87) = 0.22\). Using the standard normal distribution, we have:

\(P\left( Z > \frac{87 - 82}{\sigma} \right) = 0.22\)

This implies:

\(P\left( Z < \frac{5}{\sigma} \right) = 0.78\)

From standard normal distribution tables, \(\frac{5}{\sigma} \approx 0.772\).

Thus, \(\sigma = \frac{5}{0.772} \approx 6.48\).

(b) We need to find \(P(-4 < X - 82 < 4)\), which is equivalent to:

\(P\left( -\frac{4}{\sigma} < Z < \frac{4}{\sigma} \right)\)

Using \(\sigma = 6.48\), we have:

\(P(-0.6176 < Z < 0.6176)\)

From standard normal distribution tables, \(\Phi(0.6176) = 0.7317\).

Therefore, the probability is:

\(2 \times 0.7317 - 1 = 0.463\)