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.
$\mathrm{P}(5 \le W \le 8) = \mathrm{P}(W \le 8) - \mathrm{P}(W \le 4)$
$\therefore, \mathrm{Required ~ Probability}=0.350$