(i) The two conditions required for a binomial distribution are:

1. Constant probability of success on each trial.
2. Independent trials/events.

(ii) To find the probability that Hebe completes at least 5 puzzles in a week, we use the binomial distribution with parameters:

\(= 7 (number of trials),\)

\(= 0.7 (probability of success).\)

We need to calculate:

\(P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)\)

\(= \binom{7}{5}(0.7)^5(0.3)^2 + \binom{7}{6}(0.7)^6(0.3)^1 + (0.7)^7\)

\(= 0.647\)

(iii) To find the probability that Hebe completes 4 or fewer puzzles in exactly 3 of the 10 weeks, we first find the probability of completing 4 or fewer puzzles in a week:

\(P(X \leq 4) = 1 - P(X \geq 5) = 1 - 0.647 = 0.3529\)

Now, use the binomial distribution with parameters:

\(= 10 (weeks),\)

\(= 0.3529 (probability of completing 4 or fewer puzzles in a week).\)

We need to calculate:

\(P(Y = 3) = \binom{10}{3}(0.3529)^3(0.6471)^7\)

\(= 0.251\)