(i) The possible scores are calculated by multiplying the outcomes of the two spinners. The five-sided spinner outcomes are \(1, 1, 1, 2, 3\) and the three-sided spinner outcomes are \(1, 2, 3\). The scores are:

(ii) To find the mean, use the formula \(\mu = \sum x_i p_i\):

\(\mu = \frac{3 \times 1 + 4 \times 2 + 4 \times 3 + 1 \times 4 + 2 \times 6 + 1 \times 9}{15} = \frac{48}{15} = 3.2\)

To find the variance, use the formula \(\sigma^2 = \sum x_i^2 p_i - \mu^2\):

\(\sigma^2 = \frac{3 \times 1^2 + 4 \times 2^2 + 4 \times 3^2 + 1 \times 4^2 + 2 \times 6^2 + 1 \times 9^2}{15} - 3.2^2\)

\(\sigma^2 = \frac{224}{15} - 3.2^2 = 4.69\)

(iii) The scores greater than the mean (3.2) are 4, 6, and 9. The probability is:

\(P(X > 3.2) = \frac{1}{15} + \frac{2}{15} + \frac{1}{15} = \frac{4}{15}\)

(i) The tree diagram is as follows:

First draw two branches for the first sweet: Toffee (T) with probability \(\frac{6}{7}\) and Chocolate (C) with probability \(\frac{1}{7}\).

For the second sweet, if the first was a Toffee (T), draw branches for Toffee (T) with probability \(\frac{5}{9}\) and Chocolate (C) with probability \(\frac{4}{9}\).

If the first was a Chocolate (C), draw branches for Toffee (T) with probability \(\frac{6}{9}\) and Chocolate (C) with probability \(\frac{3}{9}\).

(ii) The probability distribution table for the number of toffees taken is:

(iii) The mean number of toffees taken is calculated as:

\(E(X) = \frac{90}{63} \times \left( \frac{10}{7} \right) = 1.43\)

(iv) The probability that the first sweet taken is a chocolate, given that the second sweet taken is a toffee, is:

\(P(\text{1st C | 2nd T}) = \frac{P(C \cap T)}{P(T)} = \frac{\frac{1}{7} \times \frac{6}{9}}{\frac{1}{7} \times \frac{6}{9} + \frac{6}{7} \times \frac{5}{9}} = \frac{6}{63} \div \frac{36}{63} = \frac{1}{6}\)

(i) The probability that Amy loses her original $1 is the probability that she fails both attempts. The probability of failing one attempt is given by:

\(P(F) = 1 - 0.2 = 0.8\)

Therefore, the probability of failing both attempts is:

\(P(\text{loses } \$1) = P(F \text{ and } F) = 0.8 \times 0.8 = 0.64\)

(ii) The probability distribution table for the amount that Amy gains is:

The probability of gaining $0.50 is the probability of failing the first attempt and succeeding on the second:

\(P(0.50) = P(F \text{ and } S) = 0.8 \times 0.2 = 0.16\)

The probability of gaining $2 is the probability of succeeding on the first attempt:

\(P(2) = 0.2\)

(iii) The expected gain is calculated as follows:

\(E(\text{winnings}) = (-1 \times 0.64) + (0.5 \times 0.16) + (2 \times 0.2)\)

\(E(\text{winnings}) = -0.64 + 0.08 + 0.4 = -0.16\)

Therefore, Amy's expected gain is -$0.16, indicating a loss of 16 cents.

(i) The probability distribution table is:

Since the sum of probabilities must equal 1, we have:

\(k + k + 4k + 9k = 15k = 1\)

Solving for \(k\), we get \(k = \frac{1}{15}\).

(ii) To find \(E(X)\), we calculate:

\(E(X) = (-1)k + 1k + 2(4k) + 3(9k) = -k + k + 8k + 27k = 35k\)

Substituting \(k = \frac{1}{15}\), we get:

\(E(X) = 35 \times \frac{1}{15} = \frac{35}{15} = \frac{7}{3}\)

To find \(Var(X)\), we use:

\(Var(X) = E(X^2) - (E(X))^2\)

\(E(X^2) = (-1)^2k + 1^2k + 2^2(4k) + 3^2(9k) = k + k + 16k + 81k = 99k\)

\(Var(X) = 99k - \left(\frac{7}{3}\right)^2\)

Substituting \(k = \frac{1}{15}\), we get:

\(Var(X) = 99 \times \frac{1}{15} - \left(\frac{7}{3}\right)^2 = \frac{99}{15} - \frac{49}{9}\)

\(Var(X) = \frac{52}{45}\)

To solve this problem, we first need to determine the possible values of \(X\) and their probabilities.

The possible outcomes for the die are \(-1, -1, 0, 0, 1, 2\) and for the spinner are \(-1, 0, 1\). The possible values of \(X\) are the sums of these outcomes.

Possible values of \(X\) are: \(-2, -1, 0, 1, 2, 3\).

Next, we calculate the expected value \(E(X)\):

\(E(X) = \frac{-4 - 4 + 0 + 4 + 4 + 3}{18} = \frac{1}{6}\)

Now, we calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{8 + 4 + 0 + 4 + 8 + 9}{18} - \left(\frac{1}{6}\right)^2\)

\(= \frac{11}{6} - \frac{1}{36}\)

\(= \frac{65}{36}\)

Thus, \(\text{Var}(X) = \frac{65}{36}\) or approximately 1.81.