(i) To find the gradient of the curve, we need to differentiate \(y = \sqrt{5x + 4}\) with respect to \(x\).

Let \(y = (5x + 4)^{1/2}\).

The derivative \(\frac{dy}{dx} = \frac{1}{2}(5x + 4)^{-1/2} \times 5\).

At \(x = 1\), \(\frac{dy}{dx} = \frac{1}{2}(5 \times 1 + 4)^{-1/2} \times 5 = \frac{5}{6}\).

(ii) We are given \(\frac{dx}{dt} = 0.03\) and need to find \(\frac{dy}{dt}\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).

Substitute \(\frac{dy}{dx} = \frac{5}{6}\) and \(\frac{dx}{dt} = 0.03\):

\(\frac{dy}{dt} = \frac{5}{6} \times 0.03 = 0.025\).

To find the \(x\)-coordinate where the \(x\)- and \(y\)-coordinates are increasing at the same rate, we need \(\frac{dy}{dx} = 1\).

First, differentiate \(y = \frac{1}{60}(3x + 1)^2\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{1}{60} \times 2(3x + 1) \times 3 = \frac{1}{10}(3x + 1)\).

Set \(\frac{dy}{dx} = 1\):

\(\frac{1}{10}(3x + 1) = 1\).

Solving for \(x\):

\(3x + 1 = 10\)

\(3x = 9\)

\(x = 3\)

Given the volume function \(V = 243 - \frac{1}{3}(9-x)^3\), we need to find \(\frac{dx}{dt}\) when \(x = 4\).

First, differentiate \(V\) with respect to \(x\):

\(\frac{dV}{dx} = (9-x)^2\).

Substitute \(x = 4\) into \(\frac{dV}{dx}\):

\(\frac{dV}{dx} = (9-4)^2 = 25\).

We know \(\frac{dV}{dt} = 3.6\) m³/hour.

Using the chain rule, \(\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}\).

So, \(3.6 = 25 \cdot \frac{dx}{dt}\).

Solving for \(\frac{dx}{dt}\):

\(\frac{dx}{dt} = \frac{3.6}{25}\) m/hour.

Convert \(\frac{dx}{dt}\) to cm/min:

\(\frac{dx}{dt} = \frac{3.6}{25} \times 100 \div 60 = 0.24\) cm/min.

Given the curve \(y = 18 - \frac{3}{8}x^{\frac{5}{2}}\), we need to find \(\frac{dy}{dt}\) when \(x = 4\) and \(\frac{dx}{dt} = 2\).

First, find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = -\frac{5}{2} \times \frac{3}{8} x^{\frac{3}{2}} = -\frac{15}{16} x^{\frac{3}{2}}\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).

Substitute \(x = 4\) and \(\frac{dx}{dt} = 2\):

\(\frac{dy}{dt} = -\frac{15}{16} \times 4^{\frac{3}{2}} \times 2\).

Calculate \(4^{\frac{3}{2}} = 8\), so:

\(\frac{dy}{dt} = -\frac{15}{16} \times 8 \times 2 = -15\).

Given the function \(f(x) = (4x + 2)^{-2}\), we need to find the derivative \(\frac{dy}{dx}\).

Using the chain rule, \(\frac{dy}{dx} = -2(4x + 2)^{-3} \times 4\).

This simplifies to \(\frac{dy}{dx} = -8(4x + 2)^{-3}\).

We know that the \(y\)-coordinate is decreasing at the rate of \(k\) and the \(x\)-coordinate is increasing at the rate of \(k\), so \(\frac{dy}{dx} = -1\).

Set \(-8(4x + 2)^{-3} = -1\).

Solving for \(x\), we get \((4x + 2)^{3} = 8\).

Taking the cube root, \(4x + 2 = 2\).

Solving for \(x\), \(4x = 0\) so \(x = 0\).

Substitute \(x = 0\) back into \(f(x)\) to find \(y\):

\(y = (4(0) + 2)^{-2} = 2^{-2} = \frac{1}{4}\).

Therefore, the coordinates of A are \((0, \frac{1}{4})\).