(i) To find the probability that each car runs out of fuel before travelling 367 km, we standardize the normal distribution for each car.

For the Zotoc car:

\(z = \frac{367 - 320}{21.6} = 2.176\)

The probability \(P(Z < 2.176)\) is approximately 0.985.

For the Ganmor car:

\(z = \frac{367 - 350}{7.5} = 2.267\)

The probability \(P(Z < 2.267)\) is approximately 0.988.

(ii) We are given that the probability that a Zotoc car can travel at least \(320 + d\) km is 0.409. This corresponds to a z-score of 0.23 (since \(P(Z > 0.23) = 0.409\)).

Standardizing, we have:

\(0.23 = \frac{x - 320}{21.6}\)

Solving for \(x\), we get:

\(x = 0.23 \times 21.6 + 320 = 324.968\)

Thus, \(d = 324.968 - 320 = 4.97\).