(i) To find the probability that a randomly chosen bar of soap weighs more than 128 grams, we standardize the variable:

\(P(X > 128) = P\left( z > \frac{128 - 125}{4.2} \right)\)

\(= P(z > 0.7143)\)

Using the standard normal distribution table, \(P(z > 0.7143) = 1 - 0.7623 = 0.238\).

(ii) We need to find the value of \(k\) such that \(P(k < X < 128) = 0.7465\).

\(P(X < 128) = 0.7465 + 0.2377 = 0.9842\)

Using the standard normal distribution table, \(z = -2.15\) corresponds to \(P(X < k)\).

\(-2.15 = \frac{k - 125}{4.2}\)

Solving for \(k\), we get \(k = 116\).

(iii) For five bars of soap, we find the probability that more than two weigh more than 128 grams:

\(P(X > 2) = P(3, 4, 5) = 1 - P(0, 1, 2)\)

Using the binomial distribution:

\(= \binom{5}{3} (0.2377)^3 (0.7623)^2 + \binom{5}{4} (0.2377)^4 (0.7623)^1 + \binom{5}{5} (0.2377)^5\)

\(= 0.07804 + 0.01216 + 0.0007588\)

\(= 0.0910\)