(a) Differentiate \(y = 2x^{\frac{1}{2}} - 1\) to find the gradient of the tangent: \(\frac{dy}{dx} = x^{-\frac{1}{2}}\). At \(x = 4\), \(\frac{dy}{dx} = \frac{1}{2}\). The gradient of the normal is \(-\frac{1}{\frac{1}{2}} = -2\). The equation of the normal through \((4, 3)\) is \(y - 3 = -2(x - 4)\), which simplifies to \(y = -2x + 11\).

(b) Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). At \(A\), \(\frac{dy}{dx} = \frac{1}{2}\) and \(\frac{dx}{dt} = 3\). Thus, \(\frac{dy}{dt} = \frac{1}{2} \times 3 = \frac{3}{2}\).

(c) The required gradient is \(-\frac{dy}{dx} = -2\). Using the chain rule, \(\frac{dx}{dt} = \frac{1}{\text{normal gradient}} \times (-5) = \frac{1}{-2} \times (-5) = \frac{5}{2}\).