All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let $X = ~\mathrm{number ~ of ~ sixes ~ out ~ of ~ 10 ~ rolls}$.
$X \sim~ \mathrm{B} (10, \frac{1}{6})$
$\mathrm{P}(X~=~3) = 0.155$

(ii)
Let $Y = ~\mathrm{number ~ of ~ sixes ~ out ~ of ~ 60 ~ rolls}$.
$Y \sim~ \mathrm{B} (60, \frac{1}{6})$
Since $n=60$ is large, $np=10 > 5, nq=50 > 5, Y \sim~ \mathrm{N} (10, \frac{50}{6})$ approximately.
$\mathrm{P}(5 \le Y \le 8) = \mathrm{P}(4.5 \le Y \le 8.5)$ by continuity correction.
$\therefore, \mathrm{Required ~ Probability}=0.273$

(iii)
Let $W = ~\mathrm{number ~ of ~ sixes ~ out ~ of ~ 60 ~ rolls ~ of ~ the ~ biased ~ die}$.
$W \sim~ \mathrm{B} (60, \frac{1}{15})$
Since $n=60$ is large, $np=4 \textless 5,~ W \sim~ \mathrm{Po} (4)$ approximately.
$\mathrm{P}(5 \le W \le 8) = \mathrm{P}(W \le 8) - \mathrm{P}(W \le 4)$
$\therefore, \mathrm{Required ~ Probability}=0.350$

Personal Comments:
Quite a standard binomial and approximation question. Some students still forget to do continuity correction or do wrong. Please go learn the correct technique to do. At the same time, you need to state the conditions to approximate the distribution here.

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