(i) We are given \(\frac{dN}{dt} = -10\) when \(t = 0\) and \(N = 1000\). The differential equation is \(\frac{dN}{dt} = ke^{-0.02t}N\). Substituting the given values, \(-10 = k \times 1 \times 1000\), we find \(k = -0.01\).

(ii) Separate variables: \(\frac{1}{N} dN = -0.01 e^{-0.02t} dt\). Integrate both sides: \(\int \frac{1}{N} dN = \int -0.01 e^{-0.02t} dt\). This gives \(\ln N = 0.5 e^{-0.02t} + C\). Using \(N = 1000\) when \(t = 0\), \(\ln 1000 = 0.5 + C\), so \(C = \ln 1000 - 0.5\). Substitute \(N = 800\) to find \(t\): \(\ln 800 = 0.5 e^{-0.02t} + \ln 1000 - 0.5\). Solving gives \(t = 29.6\).

(iii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(N \to \frac{1000}{\sqrt{e}}\).

1. The gradient of the curve is given by \(\frac{dy}{dx} = k \frac{y^2}{x}\), where \(k\) is a constant.

2. Separate variables: \(\frac{1}{y^2} dy = \frac{k}{x} dx\).

3. Integrate both sides: \(-\frac{1}{y} = k \ln x + C\), where \(C\) is the constant of integration.

4. Use the point \((1, 1)\) to find \(C\): \(-1 = k \ln 1 + C \Rightarrow C = -1\).

5. Use the point \((e, 2)\) to find \(k\): \(-\frac{1}{2} = k \ln e - 1 \Rightarrow k = \frac{1}{2}\).

6. Substitute \(k\) and \(C\) back into the integrated equation: \(-\frac{1}{y} = \frac{1}{2} \ln x - 1\).

7. Solve for \(y\): \(y = \frac{2}{2 - \ln x}\).

(i) The gradient of the tangent at \(P\) is the same as the gradient of the curve at \(P\). Since \(TN = 2\), the horizontal distance from \(T\) to \(N\) is 2. The gradient of the tangent is \(\frac{y}{2}\), so \(\frac{dy}{dx} = \frac{1}{2}y\).

(ii) To solve the differential equation \(\frac{dy}{dx} = \frac{1}{2}y\), separate variables: \(\frac{1}{y} dy = \frac{1}{2} dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int \frac{1}{2} dx\).

This gives \(\ln y = \frac{1}{2}x + C\).

Exponentiate both sides: \(y = e^{\frac{1}{2}x + C} = e^C e^{\frac{1}{2}x}\).

Let \(e^C = A\), so \(y = Ae^{\frac{1}{2}x}\).

Using the point \((4, 3)\), substitute to find \(A\): \(3 = Ae^{2}\).

Thus, \(A = \frac{3}{e^2}\).

The equation of the curve is \(y = \frac{3}{e^2} e^{\frac{1}{2}x} = 3e^{\frac{1}{2}x - 2}\).

(i) The rate of increase of the mass of A, \(\frac{dx}{dt}\), is proportional to the mass of B. Since the total mass is constant at 50, the mass of B is \(50 - x\). Therefore, \(\frac{dx}{dt} = k(50 - x)\).

(ii) To solve \(\frac{dx}{dt} = k(50 - x)\), separate variables:

\(\frac{1}{50-x} dx = k \, dt\)

Integrate both sides:

\(-\ln|50-x| = kt + C\)

Using the initial condition \(x = 0\) when \(t = 0\), we find:

\(-\ln|50| = C\)

Thus, \(-\ln(50-x) = kt - \ln 50\).

Rearrange to find:

\(\ln\left(\frac{50}{50-x}\right) = kt\)

\(\frac{50}{50-x} = e^{kt}\)

\(50-x = 50e^{-kt}\)

\(x = 50(1 - e^{-kt})\)

Using \(x = 25\) when \(t = 10\), solve for k:

\(25 = 50(1 - e^{-10k})\)

\(0.5 = 1 - e^{-10k}\)

\(e^{-10k} = 0.5\)

\(-10k = \ln(0.5)\)

\(k = -\frac{\ln(0.5)}{10} \approx 0.0693\)

Thus, \(x = 50(1 - \exp(-0.0693t))\).

(i) Given \(\frac{dy}{dt} = -0.2xy\) and \(x = \frac{10}{(1+t)^2}\), substitute for x:

\(\frac{dy}{dt} = -0.2 \cdot \frac{10}{(1+t)^2} \cdot y = -\frac{2y}{(1+t)^2}\).

Separate variables: \(\frac{dy}{y} = -\frac{2}{(1+t)^2} dt\).

Integrate both sides: \(\int \frac{dy}{y} = \int -\frac{2}{(1+t)^2} dt\).

\(\ln y = \frac{2}{1+t} + C\).

Use initial condition \(y = 100\) when \(t = 0\):

\(\ln 100 = 2 + C\).

Thus, \(C = \ln 100 - 2\).

Final solution: \(\ln y = \frac{2}{1+t} - 2 + \ln 100\).

(ii) As \(t \to \infty\), \(\frac{2}{1+t} \to 0\), so \(\ln y \to \ln 100 - 2\).

Thus, \(y \to e^{\ln 100 - 2} = \frac{100}{e^2}\).

The mass of A, \(x = \frac{10}{(1+t)^2}\), tends to zero as \(t \to \infty\).