(ii) The mean of \(X\) is given by \(E(X) = 14 \times 0.75 = 10.5\). We evaluate the probabilities for \(X = 10\) and \(X = 11\):

\(P(10) = \binom{14}{10} (0.75)^{10} (0.25)^4 = 0.220\)

\(P(11) = \binom{14}{11} (0.75)^{11} (0.25)^3 = 0.240\)

The value of \(X\) with the highest probability is 11.

(iii) We first find the probability that Sue completes more than 11 puzzles in a month:

\(P(>11) = \binom{14}{12} (0.75)^{12} (0.25)^2 + \binom{14}{13} (0.75)^{13} (0.25)^1 + (0.75)^{14} = 0.281\)

Now, we find the probability that in exactly 3 of the next 5 months, Sue completes more than 11 puzzles:

\(P(3) = \binom{5}{3} (0.281)^3 (0.7189)^2 = 0.115\)