(i) Let the random variable \(X\) represent the number of customers who rate the food as ‘good’. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.65\). We need to find \(P(2 < X < 12)\).

\(P(2 < X < 12) = 1 - P(X \leq 2)\)

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

\(P(X = 0) = (0.35)^{12}\)

\(P(X = 1) = \binom{12}{1} (0.65)^1 (0.35)^{11}\)

\(P(X = 2) = \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = (0.35)^{12} + \binom{12}{1} (0.65)^1 (0.35)^{11} + \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = 0.0065359\)

\(P(2 < X < 12) = 1 - 0.0065359 = 0.993\)

(ii) Let \(Y\) be the number of customers who rate the food as ‘poor’. \(Y\) follows a binomial distribution with parameters \(n\) and \(p = 0.13\). We need \(P(Y \geq 1) > 0.95\).

\(P(Y \geq 1) = 1 - P(Y = 0)\)

\(P(Y = 0) = (0.87)^n\)

\(1 - (0.87)^n > 0.95\)

\((0.87)^n < 0.05\)

Solving \((0.87)^n < 0.05\) gives \(n = 22\).