(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 \(V \to 3000\).