(a) Let the mean be \(\mu = 2s\). The standard normal variable \(Z\) is given by:

\(Z = \frac{X - \mu}{s} = \frac{5.2 - 2s}{s}\)

Given \(P(X > 5.2) = 0.9\), we have \(P(Z > z) = 0.9\), which implies \(P(Z < z) = 0.1\). From standard normal tables, \(z = -1.282\).

Thus, \(\frac{5.2 - 2s}{s} = -1.282\).

Solving for \(s\):

\(5.2 - 2s = -1.282s\)

\(5.2 = 0.718s\)

\(s = \frac{5.2}{0.718} \approx 7.24 \text{ or } 7.23\)

(b) The probability that a value lies between \(\mu - \sigma\) and \(\mu + \sigma\) is \(P(|Z| < 1)\).

From standard normal tables, \(P(|Z| < 1) = 0.6826\).

Thus, the expected number of observations is:

\(0.6826 \times 800 = 546.08\)

Therefore, approximately 546 observations (accept 547).