(i) Given that the mean μ is three times the standard deviation σ, we have μ = 3σ. The probability P(X < 25) = 0.648 corresponds to a standard normal variable z = 0.38. Using the standardization formula:

\(z = \frac{25 - \mu}{\sigma} = 0.38\)

Substitute \(\mu = 3\sigma\) into the equation:

\(\frac{25 - 3\sigma}{\sigma} = 0.38\)

Solving for \(\sigma\):

\(25 - 3\sigma = 0.38\sigma\)

\(25 = 3.38\sigma\)

\(\sigma = \frac{25}{3.38} \approx 7.40\)

Then, \(\mu = 3\sigma = 3 \times 7.40 = 22.2\).

(ii) The probability that a single value of X is greater than 25 is 1 - 0.648 = 0.352. We need the probability that exactly 4 out of 6 values are greater than 25. This follows a binomial distribution:

\(P(4) = \binom{6}{4} \times (0.352)^4 \times (0.648)^2\)

\(P(4) = 15 \times (0.0152) \times (0.419904)\)

\(P(4) \approx 0.0967\)