(i) To find the mean \(m\), we use the standard normal distribution. Given that 95% of times are longer than 0.9 hours, the corresponding z-value is \(z = -1.645\) (since 5% is in the left tail).

Using the standardization formula:

\(z = \frac{0.9 - m}{0.35}\)

Substitute \(z = -1.645\):

\(-1.645 = \frac{0.9 - m}{0.35}\)

Solving for \(m\):

\(0.9 - m = -1.645 \times 0.35\)

\(0.9 - m = -0.57575\)

\(m = 0.9 + 0.57575\)

\(m = 1.47575\)

Rounding to 3 significant figures, \(m = 1.48\).

(ii) We need to find the probability that none of the 4 cars takes more than 2 hours to fit. First, find the probability that a single car takes less than 2 hours.

Standardize 2 hours:

\(z = \frac{2 - 1.48}{0.35}\)

\(z = \frac{0.52}{0.35}\)

\(z = 1.4857\)

Using the standard normal distribution table, \(P(z < 1.4857) \approx 0.933\).

The probability that none of the 4 cars takes more than 2 hours is:

\((0.933)^4 \approx 0.758\)