(i) To find \(P(2 \leq X \leq 6)\), we use the geometric distribution formula. The probability of passing on the \(n\)-th attempt is \((1-p)^{n-1}p\), where \(p = 0.3\).

\(P(2 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1) = 1 - (0.7)^6 - (0.7)\)

Calculating each term:

\((0.7)^6 = 0.117649\)

\((0.7) = 0.7\)

\(P(2 \leq X \leq 6) = 1 - 0.117649 - 0.7 = 0.582351\)

Thus, \(P(2 \leq X \leq 6) \approx 0.582\).

(ii) The expected value \(E(X)\) for a geometric distribution is given by \(\frac{1}{p}\).

\(E(X) = \frac{1}{0.3} = 3.3333\)

Thus, \(E(X) = 3 \frac{1}{3}\).