(i)(a) Differentiate \(y = \frac{1}{2}(4x - 3)^{-1}\) to find \(\frac{dy}{dx} = \left[-\frac{1}{2}(4x - 3)^{-2}\right] \times 4 = -2(4x - 3)^{-2}\).

At \(x = 1\), \(m = -2\).

The gradient of the normal is \(-\frac{1}{m} = \frac{1}{2}\).

The equation of the normal is \(y - \frac{1}{2} = \frac{1}{2}(x - 1)\).

Simplifying gives \(y = \frac{1}{2}x\).

(i)(b) Substitute \(y = \frac{1}{2}x\) into the curve equation: \(\frac{1}{2}(4x - 3)^{-1} = \frac{x}{2}\).

Solving gives \(x = -\frac{1}{4}\).

(ii) Use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

At \(x = 1\), \(\frac{dy}{dx} = -2\).

\(\frac{dy}{dt} = (-2) \times (-0.3) = 0.6\).