(a) The tree diagram shows the outcomes of the first and second throws. Each branch represents a possible outcome with a probability of \(\frac{1}{6}\).

(b) To find \(P(A)\), calculate the probability of scoring 5, 6, 7, 8, or 9:

- Score 5: \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} = \frac{7}{36}\)

- Score 6: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

- Score 7: \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}\)

- Score 8: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

- Score 9: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

Adding these probabilities gives \(P(A) = \frac{12}{36} = \frac{1}{3}\).

(c) To check independence, compare \(P(A \cap B)\) and \(P(A)P(B)\):

- \(P(B) = \frac{1}{3}\), \(P(A \cap B) = \frac{6}{36} = \frac{1}{6}\)

- \(P(A)P(B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\)

Since \(\frac{1}{6} \neq \frac{1}{9}\), events A and B are not independent.

(d) To find \(P(B | A')\), use the formula:

\(P(B | A') = \frac{P(B \cap A')}{P(A')} = \frac{\frac{6}{36}}{\frac{2}{3}} = \frac{1}{4} = 0.25\)