(i) The three conditions for a binomial distribution are:

1. Constant probability of success.
2. Independent trials.
3. Fixed number of trials with only two outcomes (success or failure).

(ii) Let the random variable \(X\) represent the number of days Julie's train is late out of 9 days. \(X\) follows a binomial distribution with parameters \(n = 9\) and \(p = 0.3\).

We need to find \(P(X > 7) + P(X < 2)\).

\(P(X > 7) = P(X = 8) + P(X = 9)\)

\(P(X = 8) = \binom{9}{8} (0.3)^8 (0.7)^1\)

\(P(X = 9) = \binom{9}{9} (0.3)^9\)

\(P(X < 2) = P(X = 0) + P(X = 1)\)

\(P(X = 0) = \binom{9}{0} (0.3)^0 (0.7)^9\)

\(P(X = 1) = \binom{9}{1} (0.3)^1 (0.7)^8\)

Calculating these probabilities:

\(P(X = 8) = 9 \times (0.3)^8 \times (0.7) = 0.000015\)

\(P(X = 9) = (0.3)^9 = 0.00000019683\)

\(P(X = 0) = (0.7)^9 = 0.0403536\)

\(P(X = 1) = 9 \times (0.3) \times (0.7)^8 = 0.121060821\)

Adding these probabilities:

\(P(X > 7) + P(X < 2) = 0.000015 + 0.00000019683 + 0.0403536 + 0.121060821 = 0.196\)