(a) The sum of probabilities must equal 1: \(p + r + 0.4 + 0.15 = 1\). Simplifying gives \(p + r + 0.55 = 1\), so \(p + r = 0.45\).

Given \(E(X) = 1.1\), we have \(0 \times 0.4 + 1 \times p + 2 \times r + 3 \times 0.15 = 1.1\). Simplifying gives \(p + 2r + 0.45 = 1.1\), so \(p + 2r = 0.65\).

Solving the equations \(p + r = 0.45\) and \(p + 2r = 0.65\) simultaneously, we subtract the first from the second: \((p + 2r) - (p + r) = 0.65 - 0.45\), giving \(r = 0.2\).

Substituting \(r = 0.2\) into \(p + r = 0.45\), we get \(p + 0.2 = 0.45\), so \(p = 0.25\).

(b) The variance \(\text{Var}(X)\) is calculated as \(E(X^2) - [E(X)]^2\).

First, calculate \(E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.25 + 2^2 \times 0.2 + 3^2 \times 0.15\).

\(E(X^2) = 0 + 0.25 + 0.8 + 1.35 = 2.4\).

Then, \(\text{Var}(X) = 2.4 - (1.1)^2 = 2.4 - 1.21 = 1.19\).

(i) The sum of all probabilities must equal 1:

\(0.24 + 0.35 + 2k + k + 0.05 = 1\)

\(0.64 + 3k = 1\)

\(3k = 0.36\)

\(k = 0.12\)

(ii) The mode is the value of \(x\) with the highest probability. Here, \(P(X = 1) = 0.35\) is the highest, so the mode is 1.

(iii) Calculate the mean:

\(\text{mean} = 1 \times 0.35 + 2 \times 0.24 + 3 \times 0.12 + 4 \times 0.05\)

\(\text{mean} = 0.35 + 0.48 + 0.36 + 0.20 = 1.39\)

Find \(P(X > 1.39) = P(X = 2) + P(X = 3) + P(X = 4)\)

\(P(X = 2) = 2k = 0.24\)

\(P(X = 3) = k = 0.12\)

\(P(X = 4) = 0.05\)

\(P(X > 1.39) = 0.24 + 0.12 + 0.05 = 0.41\)

We have the following equations based on the given information:

1. The sum of probabilities: \(p + q + r + 0.4 = 1\)

2. The expected value: \(-3p + 2r + 4 \times 0.4 = 2.3\)

3. The variance: \((-3)^2p + 2^2r + 4^2 \times 0.4 - 2.3^2 = 3.01\)

From equation 1, we have:

\(p + q + r = 0.6\)

From equation 2, we simplify:

\(-3p + 2r = 0.7\)

From equation 3, we simplify:

\(9p + 4r = 1.9\)

Solving \(-3p + 2r = 0.7\) and \(9p + 4r = 1.9\) simultaneously:

Multiply the first equation by 2:

\(-6p + 4r = 1.4\)

Subtract from the second equation:

\(9p + 6p = 1.9 - 1.4\)

\(15p = 0.5\)

\(p = \frac{1}{30}\)

Substitute \(p\) back into \(-3p + 2r = 0.7\):

\(-3 \times \frac{1}{30} + 2r = 0.7\)

\(-0.1 + 2r = 0.7\)

\(2r = 0.8\)

\(r = \frac{2}{5}\)

Substitute \(p\) and \(r\) into \(p + q + r = 0.6\):

\(\frac{1}{30} + q + \frac{2}{5} = 0.6\)

\(q = 0.6 - 0.4 - 0.0333\)

\(q = \frac{1}{6}\)

(i) To find the value of p, we use the fact that the total probability must sum to 1:

\(0.3 + 0.15 + 3p + 2p + 0.05 = 1\)

\(0.5 + 5p = 1\)

\(5p = 0.5\)

\(p = 0.1\)

(ii) (a) To find the probability that the sum of the scores is 4, we consider the possible combinations:

\(P(X = 1, Y = 3) = 0.3 imes 0.2 = 0.06\)

\(P(X = 2, Y = 2) = 0.15 imes 0.5 = 0.075\)

\(P(X = 3, Y = 1) = 0.3 imes 0.3 = 0.09\)

\(P( ext{sum is 4}) = 0.06 + 0.075 + 0.09 = 0.225\)

(ii) (b) To find the probability that the product of the scores is less than 8, we consider the possible combinations:

\(P(X = 1, Y = ext{anything}) = 0.3\)

\(P(X = 2, Y = ext{anything}) = 0.15\)

\(P(X = 3, Y = 1, 2) = 0.3 imes 0.8 = 0.24\)

\(P(X = 4, Y = 1) = 0.2 imes 0.3 = 0.06\)

\(P(X = 5, Y = 1) = 0.05 imes 0.3 = 0.015\)

\(P( ext{product} < 8) = 0.3 + 0.15 + 0.24 + 0.06 + 0.015 = 0.765\)

(i) The probability of any other number is \(\frac{9}{70}\). Therefore, \(P(X < 2) = \frac{27}{70} + \frac{1}{10} = \frac{34}{70} = \frac{17}{35}\).

(ii) Calculate the expected value \(E(X) = \frac{108}{70} = \frac{54}{35} \approx 1.543\). The variance is given by:

\(\text{Var}(X) = ((-2)^2 + \ldots + 5^2) \times \frac{9}{70} - \left(\frac{54}{35}\right)^2\)

\(= 5.33\).

(iii) For \(P(-a \leq X \leq 2a) = \frac{17}{35}\), we find \(a = 1\).