(N/A) When the blue die shows $1$,the grey die could show any one of the numbers $1, 2, 3, 4, 5, 6$. The same is true when the blue die shows $2, 3, 4, 5,$ or $6$. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.
Note that the pair $(1, 4)$ is different from $(4, 1)$.
So,the total number of possible outcomes $= 6 \times 6 = 36$.
$(i)$ The outcomes favourable to the event "the sum of the two numbers is $8$" denoted by $E$,are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
i.e.,the number of outcomes favourable to $E = 5$.
Hence,$P(E) = \frac{5}{36}$.
$(ii)$ As you can see,there is no outcome favourable to the event $F$,"the sum of two numbers is $13$".
So,$P(F) = \frac{0}{36} = 0$.
$(iii)$ As you can see,all the outcomes are favourable to the event $G$,"sum of two numbers $\leq 12$".
So,$P(G) = \frac{36}{36} = 1$.