(a) $n(S)=6 \times 6=36$
Let $A$ represent obtaining a sum greater than 9 and $B$ represents black die resulted in a 5 .
$\mathrm{A}=\{46,64,55,36,63,45,54,65,56,66\}$
$$
n(\mathrm{~A})=10 \Rightarrow P(A)=\frac{n(A)}{n(S)}=\frac{10}{216}
$$
$$
\begin{aligned}
&\mathrm{B}=\{51,52,53,54,55,56\} \Rightarrow n(\mathrm{~B})=6 \\
&\mathrm{P}(\mathrm{B})=\frac{6}{216}, \\
&A \cap B=\{55,56\} \Rightarrow n(A \cap B)=2 \\
&P(A \cap B)=\frac{2}{216}, \\
&P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{2}{216}}{\frac{6}{216}}=\frac{2}{6}=\frac{1}{3} .
\end{aligned}
$$
(b) Let $\mathrm{A}$ denotes the sum is 8
$\therefore \quad \mathrm{A}=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$
$\mathrm{B}=$ Red die results in a number less than 4 either first or second die is red.
$\mathrm{B}=\{(1,1),(1,2)(1,3)(1,4),(1,5),(1,6)$,
$(2,1)(2,2)(2,3),(2,4),(2,5),(2,6),(3,1)$,
$(3,2)(3,3),(3,4),(3,5),(3,6)\}$
$A \cap B=\{(2,6),(3,5)\}$
$$
\therefore \quad P(A \cap B)=\frac{2}{36}=\frac{1}{18}, P(B)=\frac{18}{36}=\frac{1}{2}
$$
Hence $P(A \mid B)=\frac{P(A B)}{P(B)}=\frac{1}{9}$.