8. In a certain World Cup, the Brazilian team has a probability of $\frac{1}{2}$ to win against each opponent, a probability of $\frac{1}{3}$ to draw, and a probability of $\frac{1}{6}$ to lose. Winning a match earns 3 points, drawing earns 1 point, and losing earns 0 points. It is known that each team plays three matches in the group stage, and a team needs 4 points or more to advance from the group stage. In the knockout stage, if the teams draw, they go to a penalty shootout. It is known that Brazil's probability of winning a penalty shootout is $\frac{3}{5}$. What is the probability that Brazil will win the championship and have exactly one penalty shootout in the four knockout matches?
8. $\frac{1}{12}$.

Let $\xi$ be the points Brazil scores in the group stage. Then $\xi=0,1, \cdots, 7,9$.
From the problem, we have
$$
\begin{array}{l}
P(\xi=4)=C_{3}^{1} \times \frac{1}{2} C_{2}^{1} \times \frac{1}{3} C_{1}^{1} \times \frac{1}{6}=\frac{1}{6}, \\
P(\xi=5)=C_{3}^{2}\left(\frac{1}{3}\right)^{2} \times \frac{1}{2}=\frac{1}{6}, \\
P(\xi=6)=C_{3}^{2}\left(\frac{1}{2}\right)^{2} \times \frac{1}{6}=\frac{1}{8}, \\
P(\xi=7)=\mathrm{C}_{3}^{2}\left(\frac{1}{2}\right)^{2} \times \frac{1}{3}=\frac{1}{4}, \\
P(\xi=9)=\left(\frac{1}{2}\right)^{3}=\frac{1}{8} .
\end{array}
$$

Let "Brazil advancing from the group stage" be event $A$. Then
$$
\begin{array}{l}
P(A)=P(\xi \geqslant 4) \\
=\frac{1}{6}+\frac{1}{6}+\frac{1}{8}+\frac{1}{4}+\frac{1}{8}=\frac{5}{6} .
\end{array}
$$

Let "Brazil winning the championship and winning exactly one penalty shootout in the four knockout matches" be event $B$. Then
$$
P(B)=\frac{5}{6} \mathrm{C}_{4}^{1}\left(\frac{1}{2}\right)^{3} \times \frac{1}{3} \times \frac{3}{5}=\frac{1}{12} .
$$