(i) We have two conditions:

1. 8 out of 10 children can jump more than 127 cm, so \(P(X > 127) = 0.8\). This corresponds to a z-score of \(-0.842\).

2. 1 out of 3 children can jump more than 135 cm, so \(P(X > 135) = 0.333\). This corresponds to a z-score of \(0.431\).

Using the z-score formula \(z = \frac{x - \mu}{\sigma}\), we set up the equations:

\(0.431 = \frac{135 - \mu}{\sigma}\)

\(-0.842 = \frac{127 - \mu}{\sigma}\)

Solving these equations simultaneously gives \(\mu = 132\) and \(\sigma = 6.29\).

(ii) To find the probability that a child will not be able to jump 145 cm, we calculate \(P(X < 145)\).

Standardize: \(z = \frac{145 - 132}{6.29} = 2.023\).

Using normal distribution tables, \(P(z < 2.023) = 0.978\).

(iii) Let \(p = \frac{1}{3}\) be the probability that a child can jump more than 135 cm.

We need \(P(\text{at least 2 out of 8}) = 1 - P(0, 1)\).

\(P(0, 1) = \left(\frac{2}{3}\right)^8 + \binom{8}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^7\).

Calculate: \(P(0, 1) = 0.195\).

Thus, \(P(\text{at least 2}) = 1 - 0.195 = 0.805\).