(i) We start with the given \(V = \frac{4}{3}h^3\). Differentiating with respect to \(t\), we have \(\frac{dV}{dt} = 4h^2 \frac{dh}{dt}\). The rate of liquid poured in is 20 m\(^3\) per hour, and the rate of leakage is \(kh^2\). Thus, \(\frac{dV}{dt} = 20 - kh^2\). Equating the two expressions for \(\frac{dV}{dt}\), we have:

\(4h^2 \frac{dh}{dt} = 20 - kh^2\)

When \(h = 1\), \(\frac{dh}{dt} = 4.95\), so:

\(4(1)^2(4.95) = 20 - k(1)^2\)

\(19.8 = 20 - k\)

\(k = 0.2\)

Substituting \(k = 0.2\) back, we get:

\(4h^2 \frac{dh}{dt} = 20 - 0.2h^2\)

\(\frac{dh}{dt} = \frac{5}{h^2} - \frac{1}{20}\)

(ii) We need to verify:

\(\frac{20h^2}{100-h^2} = -20 + \frac{2000}{(10-h)(10+h)}\)

Expanding the right side:

\(\frac{2000}{(10-h)(10+h)} = \frac{2000}{100-h^2}\)

Thus, \(-20 + \frac{2000}{100-h^2} = \frac{2000 - 20(100-h^2)}{100-h^2} = \frac{20h^2}{100-h^2}\)

(iii) Separating variables in the differential equation:

\(\frac{dh}{\frac{5}{h^2} - \frac{1}{20}} = dt\)

Integrating both sides, we have:

\(\int \frac{h^2}{5 - \frac{h^2}{20}} \, dh = \int dt\)

\(\int \frac{20h^2}{100-h^2} \, dh = t + C\)

Using partial fraction decomposition:

\(\frac{20h^2}{100-h^2} = -20 + \frac{2000}{(10-h)(10+h)}\)

Integrating gives:

\(-20h + 100 \ln(10+h) + 100 \ln(10-h) = t + C\)

\(t = 100 \ln \left( \frac{10+h}{10-h} \right) + C\)

(i) The gradient of the curve at \(P\) is given by \(\frac{PN}{TN} = \frac{y}{TN}\). The area of triangle \(PTN\) is \(\frac{1}{2} \times TN \times y = \tan x\). Therefore, \(TN = \frac{2 \tan x}{y}\). The gradient \(\frac{dy}{dx} = \frac{y}{TN} = \frac{y^2}{2 \tan x} = \frac{1}{2}y^2 \cot x\).

(ii) Separate variables: \(\frac{dy}{y^2} = \frac{1}{2} \cot x \, dx\). Integrate both sides: \(-\frac{1}{y} = \frac{1}{2} \ln(\sin x) + C\). Given \(y = 2\) when \(x = \frac{1}{6}\pi\), substitute to find \(C\): \(-\frac{1}{2} = \frac{1}{2} \ln(\sin \frac{1}{6}\pi) + C\). Solve for \(C\) and substitute back to find \(y\): \(y = \frac{2}{1 - \ln(2 \sin x)}\).

(i) The rate of increase in height is proportional to \((9 - h)^{\frac{1}{3}}\), so we can write \(\frac{dh}{dt} = k(9 - h)^{\frac{1}{3}}\). Given \(\frac{dh}{dt} = 0.2\) when \(h = 1\), we have \(0.2 = k(9 - 1)^{\frac{1}{3}}\). Solving for \(k\), we find \(k = 0.1\). Thus, \(\frac{dh}{dt} = 0.1(9 - h)^{\frac{1}{3}}\).

(ii) Separate variables: \(\frac{dh}{(9 - h)^{\frac{1}{3}}} = 0.1 \, dt\). Integrate both sides: \(\int (9 - h)^{-\frac{1}{3}} \, dh = \int 0.1 \, dt\). This gives \(-\frac{3}{2}(9 - h)^{\frac{2}{3}} = 0.1t + C\). Use initial conditions \(t = 0, h = 1\) to find \(C\): \(-\frac{3}{2}(9 - 1)^{\frac{2}{3}} = C\). Solve for \(h\): \(h = 9 - \left(4 - \frac{1}{15}t\right)^{\frac{3}{2}}\).

(iii) The maximum height occurs when \(\frac{dh}{dt} = 0\), which implies \(9 - h = 0\), so \(h = 9\). The time taken is when \(h = 9\), which is \(t = 60\) years.

(iv) For half the maximum height, \(h = \frac{9}{2}\). Substitute into the expression for \(h\): \(\frac{9}{2} = 9 - \left(4 - \frac{1}{15}t\right)^{\frac{3}{2}}\). Solve for \(t\) to find \(t \approx 19.1\) years.

(i) The rate of formation of the substance is proportional to its mass, so we can write \(\frac{dx}{dt} = kx\) for some constant \(k\). The substance is also being removed at a constant rate of 25 grams per minute, so \(\frac{dx}{dt} = kx - 25\). Given \(\frac{dx}{dt} = 75\) when \(x = 1000\), we have:

\(75 = k(1000) - 25\)

\(75 = 1000k - 25\)

\(100 = 1000k\)

\(k = 0.1\)

Thus, \(\frac{dx}{dt} = 0.1x - 25\) can be rewritten as \(\frac{dx}{dt} = 0.1(x - 250)\).

(ii) To solve the differential equation \(\frac{dx}{dt} = 0.1(x - 250)\), we separate variables:

\(\frac{1}{x - 250} dx = 0.1 dt\)

Integrate both sides:

\(\int \frac{1}{x - 250} dx = \int 0.1 dt\)

\(\ln|x - 250| = 0.1t + C\)

Exponentiate both sides:

\(x - 250 = e^{0.1t + C}\)

\(x - 250 = e^C e^{0.1t}\)

Let \(e^C = A\), then \(x = A e^{0.1t} + 250\).

Using the initial condition \(x = 1000\) when \(t = 0\):

\(1000 = A e^{0} + 250\)

\(1000 = A + 250\)

\(A = 750\)

Thus, \(x = 750 e^{0.1t} + 250\).

Rearranging gives \(x = 250(3e^{0.1t} + 1)\).

(i) The rate of change of volume \(\frac{dV}{dt}\) is given by the inflow minus the outflow: \(\frac{dV}{dt} = 30 - k\sqrt{h}\). Since \(V = 1000h\), \(\frac{dV}{dt} = 1000 \frac{dh}{dt}\). Equating gives \(1000 \frac{dh}{dt} = 30 - k\sqrt{h}\). Given \(\frac{dh}{dt} = 0.02\) when \(h = 1\), we find \(k = 10\). Thus, \(\frac{dh}{dt} = 0.01(3 - \sqrt{h})\).

(ii) Substitute \(x = 3 - \sqrt{h}\), then \(\sqrt{h} = 3 - x\) and \(h = (3 - x)^2\). Differentiate to get \(\frac{dh}{dt} = -2(3-x) \frac{dx}{dt}\). Substitute into the differential equation to get \(-2(3-x) \frac{dx}{dt} = 0.01(3 - (3-x))\), simplifying to \((x-3) \frac{dx}{dt} = 0.005x\). Separate variables and integrate: \(\int \frac{x-3}{x} \, dx = \int 0.005 \, dt\). This gives \(x - 3 \ln x = 0.005t + C\). Using \(x = 3\) when \(t = 0\), find \(C = 0\). Thus, \(t = 200(x - 3 \ln x + 3 \ln 3)\).

(iii) When \(h = 4\), \(\sqrt{h} = 2\), so \(x = 3 - 2 = 1\). Substitute \(x = 1\) into the expression for \(t\): \(t = 200(1 - 3 \ln 1 + 3 \ln 3) = 200(1 + 3 \ln 3)\). Calculate to find \(t = 259\) seconds.