(i) To find the number of days the sales exceed 3900 litres, we standardize the value:

\(P(x > 3900) = P\left( z > \frac{3900 - 4520}{560} \right) = P(z > -1.107) = \Phi(1.107)\)

Using the standard normal distribution table, \(\Phi(1.107) = 0.8657\).

The expected number of days is \(365 \times 0.8657 = 315.98\), so approximately 315 or 316 days.

(ii) Given \(P(X > 8000) = 0.122\), we find the z-score:

\(z = 1.165\)

Using the standardization formula:

\(1.165 = \frac{8000 - m}{560}\)

Solving for \(m\):

\(m = 8000 - 1.165 \times 560 = 7350\)

(iii) The probability that sales exceed 8000 litres on fewer than 2 of 6 days is calculated using the binomial distribution:

\(P(0, 1) = (0.878)^6 + \binom{6}{1}(0.122)^1(0.878)^5\)

Calculating gives \(0.840\), which can be accepted as 0.84.