(i) The sum of probabilities must equal 1: \(0.2 + 0.1 + p + 0.1 + q = 1\). Simplifying gives \(p + q = 0.6\).

Given \(E(X) = 1.7\), we have \(\Sigma px = -0.4 + 0 + p + 0.3 + 4q = 1.7\). Simplifying gives \(p + 4q = 1.8\).

Solving the simultaneous equations \(p + q = 0.6\) and \(p + 4q = 1.8\), we find \(p = 0.2\) and \(q = 0.4\).

(ii) The variance is given by \(\text{Var}(X) = \Sigma x^2p - (E(X))^2\).

Calculate \(\Sigma x^2p = 4 \times 0.2 + 1 \times p + 9 \times 0.1 + 16 \times q\).

Substitute \(p = 0.2\) and \(q = 0.4\) to get \(4 \times 0.2 + 1 \times 0.2 + 9 \times 0.1 + 16 \times 0.4 = 8.3\).

Then \(\text{Var}(X) = 8.3 - 1.7^2 = 8.3 - 2.89 = 5.41\).

To find the values of p and q, we use the following conditions:

1. The sum of all probabilities must equal 1: \(0.15 + p + 0.4 + q = 1\).
2. The expected value is given by \(E(X) = 3.05\), which means \(1 \times 0.15 + 2 \times p + 3 \times 0.4 + 6 \times q = 3.05\).

From the first condition, we have:

\(p + q = 0.45\).

From the second condition, we have:

\(0.15 + 2p + 1.2 + 6q = 3.05\).

Simplifying the second equation:

\(2p + 6q = 3.05 - 0.15 - 1.2 = 1.7\).

Now, solve the simultaneous equations:

1. \(p + q = 0.45\)
2. \(2p + 6q = 1.7\)

From equation (1), express \(p\) in terms of \(q\):

\(p = 0.45 - q\).

Substitute into equation (2):

\(2(0.45 - q) + 6q = 1.7\).

\(0.9 - 2q + 6q = 1.7\).

\(0.9 + 4q = 1.7\).

\(4q = 1.7 - 0.9 = 0.8\).

\(q = 0.2\).

Substitute \(q = 0.2\) back into \(p = 0.45 - q\):

\(p = 0.45 - 0.2 = 0.25\).

Thus, the values are \(p = 0.25\) and \(q = 0.2\).