(i) Since the sum of all probabilities must equal 1, we have:

\(0.3 + a + b + 0.25 = 1\)

\(a + b = 0.45\)

(ii) The expected value \(E(X)\) is given by:

\(E(X) = 1 \times 0.3 + 3 \times a + 5 \times b + 7 \times 0.25\)

Given \(E(X) = 4\), we have:

\(0.3 + 3a + 5b + 1.75 = 4\)

\(3a + 5b = 1.95\)

We now solve the system of equations:

1. \(a + b = 0.45\)

2. \(3a + 5b = 1.95\)

Substitute \(a = 0.45 - b\) into the second equation:

\(3(0.45 - b) + 5b = 1.95\)

\(1.35 - 3b + 5b = 1.95\)

\(2b = 0.6\)

\(b = 0.3\)

Substitute \(b = 0.3\) back into \(a + b = 0.45\):

\(a + 0.3 = 0.45\)

\(a = 0.15\)

Thus, \(a = 0.15\) and \(b = 0.3\).