(i) We know that \(P(X > 410) = \frac{225}{6000} = 0.0375\). Therefore, \(P\left( Z > \frac{410 - 400}{\sigma} \right) = 0.0375\), which implies \(P(Z > z) = 0.0375\). Thus, \(P(Z < z) = 0.9625\).

The corresponding \(z\) value for 0.9625 is approximately 1.78. Therefore, \(\frac{10}{\sigma} = 1.78\), giving \(\sigma = \frac{10}{1.78} = 5.62\) grams.

(ii) We need \(P(Z < -1.5)\) and \(P(Z > 1.5)\). This is \(\Phi(-1.5) + 1 - \Phi(1.5) = 2 - 2\Phi(1.5)\).

Using \(\Phi(1.5) \approx 0.9332\), we have \(2 - 2 \times 0.9332 = 0.1336\).

The expected number of packets is \(500 \times 0.1336 = 66.8\), so approximately 66 or 67 packets.