(i) Given \(y = 2^x\), we have \(2^{-x} = \frac{1}{y}\). Substitute into the equation:

\(2^x - 2^{-x} = y - \frac{1}{y} = 1\).

Multiply through by \(y\) to clear the fraction:

\(y^2 - 1 = y\).

Rearrange to form a quadratic equation:

\(y^2 - y - 1 = 0\).

(ii) Solve the quadratic equation \(y^2 - y - 1 = 0\) using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -1, c = -1\).

\(y = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\).

Since \(y = 2^x\) and \(y > 0\), choose the positive root:

\(y = \frac{1 + \sqrt{5}}{2}\).

Now solve \(2^x = \frac{1 + \sqrt{5}}{2}\) by taking logarithms:

\(x = \log_2 \left( \frac{1 + \sqrt{5}}{2} \right)\).

Calculate \(x \approx 0.694\).