When two dice are thrown there are $6 \times 6=6^2$ possible outcomes and
$\begin{aligned} S=&\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ &(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ &(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ &(4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\ &(5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned}$
$\therefore A:$ the sum is greater than 8
$\begin{array}{r}
=\{(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4), \\
(5,6),(6,5),(6,6)\}
\end{array}$
$B: 2$ occurs on either die $=\{(1,2),(2,2),(3,2),(4,2),(5,2),(6,2)$,
$(2,1),(2,3),(2,4),(2,5),(2,6)\}$
$C$ : The sum is at least 7 and a multiple of 3 $=\{(3,6),(6,3),(5,4),(4,5),(6,6)\}$
$A \cap B=\phi, B \cap C=\phi$
and $A \cap C=\{(3,6),(6,3),(5,4),(4,5),(6,6)\}$
(i) Since $A \cap B=\phi$, So, $A$ and $B$ are mutually exclusive.
(ii) Since $B \cap C=\phi$, So, $B$ and $C$ are mutually exclusive.
(iii) Since $A \cap C=\phi$, So, $A$ and $C$ are not mutually exclusive.