(i) To find the probability that a journey takes between 85 and 100 minutes, we standardize the values using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For \(x = 85\), \(z = \frac{85 - 100}{7} = -2.143\).

The probability \(P(85 < x < 100) = 0.5 - P(z < -2.143)\).

Using the standard normal distribution table, \(P(z < -2.143) = 1 - \Phi(2.143) = 0.0161\).

Thus, \(P(85 < x < 100) = 0.5 - 0.0161 = 0.484\).

(ii) For 'standard' journeys, which are the middle 34%, we find the z-scores corresponding to the cumulative probabilities of 0.33 and 0.67.

Using \(z = \Phi^{-1}(0.67) = 0.44\).

For the upper limit, \(0.44 = \frac{a - 100}{7}\), solving gives \(a = 103.1\) minutes.

For the lower limit, \(-0.44 = \frac{a - 100}{7}\), solving gives \(a = 96.9\) minutes.