(i) Let the probability of choosing a chocolate biscuit be \(p = \frac{1}{4}\) and a cream biscuit be \(1 - p = \frac{3}{4}\). The probability of having 5 chocolate and 5 cream biscuits in a box of 10 is given by:

\(P(\text{equal}) = \left(0.25\right)^5 \times \left(0.75\right)^5 \times \binom{10}{5} = 0.0584\)

(ii) The probability that exactly 1 out of 8 boxes contains equal numbers of cream and chocolate biscuits is:

\((0.0584)^1 \times (0.9416)^7 \times \binom{8}{1} = 0.307\)

(iii) For a large box of 120 biscuits, use the normal approximation. Let \(X\) be the number of chocolate biscuits. Then \(\mu = 120 \times 0.25 = 30\) and \(\sigma^2 = 30 \times 0.75 = 22.5\).

Using continuity correction, find \(P(X < 35)\):

\(P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{22.5}}\right) = \Phi(0.949)\)

From standard normal tables, \(\Phi(0.949) = 0.829\).