(i) To find \(\frac{dy}{dx}\), use the quotient rule. Let \(u = 12\) and \(v = 3 - 2x\). Then \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = -2\).

Using the quotient rule: \(\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} = \frac{(3 - 2x) \cdot 0 - 12 \cdot (-2)}{(3 - 2x)^2} = \frac{24}{(3 - 2x)^2}\).

(ii) Given \(\frac{dy}{dt} = 0.4\) and \(\frac{dx}{dt} = 0.15\), use the chain rule: \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{0.4}{0.15} = \frac{8}{3}\).

Equate this to the derivative found in part (i): \(\frac{24}{(3 - 2x)^2} = \frac{8}{3}\).

Cross-multiply to solve for \(x\): \(24 \cdot 3 = 8 \cdot (3 - 2x)^2\).

\(72 = 8(3 - 2x)^2\)

\(9 = (3 - 2x)^2\)

Taking the square root: \(3 - 2x = 3\) or \(3 - 2x = -3\).

Solving these gives \(x = 0\) or \(x = 3\).

Given the curve \(y = \frac{8}{x} + 2x\), we need to find \(\frac{dy}{dt}\) when \(x = 1\) and \(\frac{dx}{dt} = 0.04\).

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

\(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{8}{x} + 2x \right) = -\frac{8}{x^2} + 2\).

At point \(A\) where \(x = 1\), substitute \(x = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = -\frac{8}{1^2} + 2 = -8 + 2 = -6\).

Now, use the chain rule to find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).

Substitute \(\frac{dy}{dx} = -6\) and \(\frac{dx}{dt} = 0.04\):

\(\frac{dy}{dt} = -6 \times 0.04 = -0.24\).

(a) To find the rate at which \(h\) is increasing, we use the chain rule. First, differentiate \(V\) with respect to \(h\):

\(\frac{dV}{dh} = \frac{4}{3} \times 3(25 + h)^2 = 4(25 + h)^2\).

At \(h = 10\), \(\frac{dV}{dh} = 4(25 + 10)^2 = 4900\).

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

Given \(\frac{dV}{dt} = 500\), we have:

\(500 = 4900 \cdot \frac{dh}{dt}\).

Solving for \(\frac{dh}{dt}\), we get \(\frac{dh}{dt} = \frac{500}{4900} = 0.102 \text{ cm/s}\).

(b) Given \(\frac{dh}{dt} = 0.075 \text{ cm/s}\), use the chain rule:

\(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = 4(25 + h)^2 \cdot 0.075\).

Set \(\frac{dV}{dt} = 500\):

\(500 = 4(25 + h)^2 \cdot 0.075\).

Solving for \(h\), we have:

\((25 + h)^2 = \frac{5000}{3}\).

\(25 + h = \sqrt{\frac{5000}{3}}\).

\(h = \sqrt{\frac{5000}{3}} - 25 \approx 15.82\).

Substitute \(h\) back into the volume formula:

\(V = \frac{4}{3}(25 + 15.82)^3 - \frac{62500}{3}\).

\(V \approx 69900 \text{ cm}^3\).

The area \(A\) of a circle is given by \(A = \pi r^2\), where \(r\) is the radius.

To find the rate at which the area is increasing, we need \(\frac{dA}{dt}\).

Using the chain rule, \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\).

First, find \(\frac{dA}{dr}\):

\(\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r\).

Given \(\frac{dr}{dt} = 3\) m/h and \(r = 50\) m at midday, substitute these values:

\(\frac{dA}{dt} = 2\pi (50) \times 3 = 300\pi\).

Thus, \(\frac{dA}{dt} = 300\pi\) square metres per hour, or approximately 942 square metres per hour.

(i) To find \(\frac{dy}{dx}\), we differentiate each term of \(y = 4\sqrt{x} + \frac{2}{\sqrt{x}}\).

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

Differentiate: \(\frac{dy}{dx} = 4 \cdot \frac{1}{2}x^{-\frac{1}{2}} + 2 \cdot (-\frac{1}{2})x^{-\frac{3}{2}}\).

Simplify: \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - x^{-\frac{3}{2}}\).

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

Given \(\frac{dx}{dt} = 0.12\) and \(x = 4\), substitute into \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - x^{-\frac{3}{2}}\).

At \(x = 4\), \(\frac{dy}{dx} = 2(4)^{-\frac{1}{2}} - (4)^{-\frac{3}{2}} = \frac{1}{\sqrt{4}} - \frac{1}{8} = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}\).

Thus, \(\frac{dy}{dt} = \frac{3}{8} \times 0.12 = 0.105\).