(a) The rate of water being added is 30 litres per minute, so \(a = 30\). The rate of water being lost is proportional to the volume, \(0.01V\), so \(b = 0.01\).

(b) The differential equation is \(\frac{dV}{dt} = 30 - 0.01V\). Separate variables and integrate:

\(\int \frac{1}{30 - 0.01V} \, dV = \int \, dt\)

\(-100 \ln(30 - 0.01V) = t + C\)

Using initial condition \(V = 0\) when \(t = 0\), find \(C\):

\(-100 \ln(30) = C\)

Thus, \(-100 \ln(30 - 0.01V) = t - 100 \ln(30)\)

\(100 \ln \left( \frac{30}{30 - 0.01V} \right) = t\)

For \(V = 1000\):

\(100 \ln \left( \frac{30}{30 - 10} \right) = t\)

\(t = 40.5\)

(c) Solving for \(V\) in terms of \(t\):

\(V = 3000(1 - e^{-0.01t})\)

As \(t \to \infty\), \(e^{-0.01t} \to 0\), so \(V \to 3000\).

(i) To find an expression for \(N\) in terms of \(t\), we start by separating variables and integrating both sides:

\(\int \frac{1}{N^{0.5}} \, dN = \int 1.2e^{-0.02t} \, dt\).

This gives:

\(2N^{0.5} = -60e^{-0.02t} + C\).

Using the initial condition \(N = 100\) when \(t = 0\), we find \(C\):

\(2(100)^{0.5} = -60e^{0} + C\)

\(20 = -60 + C\)

\(C = 80\).

Thus, the equation becomes:

\(2N^{0.5} = -60e^{-0.02t} + 80\).

Solving for \(N\):

\(N^{0.5} = -30e^{-0.02t} + 40\)

\(N = (40 - 30e^{-0.02t})^2\).

(ii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(N \to 40^2 = 1600\).

(i) To solve the differential equation \(\frac{dN}{dt} = \frac{N(1800 - N)}{3600}\), we first separate variables:

\(\frac{1}{N(1800 - N)} = \frac{1}{3600} \frac{dN}{dt}.\)

We use partial fraction decomposition:

\(\frac{1}{N(1800 - N)} = \frac{A}{N} + \frac{B}{1800 - N}.\)

Solving for \(A\) and \(B\), we find:

\(\frac{2}{N} + \frac{2}{1800 - N}.\)

Integrating both sides gives:

\(2 \ln N - 2 \ln (1800 - N) = \frac{t}{2} + C.\)

Using the initial condition \(N = 300\) when \(t = 0\), we solve for \(C\):

\(2 \ln 300 - 2 \ln 1500 = C.\)

Thus, the solution is:

\(N = \frac{1800e^{t/2}}{5 + e^{t/2}}.\)

(ii) As \(t \to \infty\), \(e^{t/2} \to \infty\), so \(N \to 1800\).

(i) Separate variables: \(\frac{1}{\theta - A} d\theta = -k \, dt\).

Integrate both sides: \(\ln(\theta - A) = -kt + C\).

Use initial condition \(\theta = 4A\) when \(t = 0\):

\(\ln(4A - A) = C \Rightarrow C = \ln 3A\).

Thus, \(\ln(\theta - A) = -kt + \ln 3A\).

(ii) Substitute \(\theta = 3A\), \(t = 1\):

\(\ln(3A - A) = -k(1) + \ln 3A\).

\(\ln 2A = -k + \ln 3A\).

\(k = \ln 3A - \ln 2A = \ln \frac{3}{2}\).

(iii) Substitute \(t = 2\) into \(\ln(\theta - A) = -kt + \ln 3A\):

\(\ln(\theta - A) = -2\ln \frac{3}{2} + \ln 3A\).

\(\ln(\theta - A) = \ln \left(\frac{3A}{\left(\frac{3}{2}\right)^2}\right)\).

\(\theta - A = \frac{3A}{\frac{9}{4}} = \frac{4}{3}A\).

\(\theta = \frac{4}{3}A + A = \frac{7}{3}A\).

(i) Separate variables and integrate:

\(\int \frac{1}{N} \, dN = \int k \cos(0.02t) \, dt\)

\(\ln N = \frac{k}{0.02} \sin(0.02t) + C\)

Using \(t = 0, N = 125\):

\(\ln 125 = C\)

Thus, \(\ln N = \frac{k}{0.02} \sin(0.02t) + \ln 125\)

(ii) Given \(N = 166\) when \(t = 30\):

\(\ln 166 = \frac{k}{0.02} \sin(0.6) + \ln 125\)

Solving for \(k\), \(k \approx 0.0100479\) (accept \(k = 0.01\))

(iii) Expression for \(N\):

\(N = 125e^{0.502 \sin(0.02t)}\)

Least value occurs when \(\sin(0.02t) = -1\):

\(N = 125e^{-0.502} \approx 75.6\)