(i) The rate of increase of N is proportional to (N - 150), so we write the differential equation as \(\frac{dN}{dt} = k(N - 150)\).

(ii) Given \(\frac{dN}{dt} = 60\) and \(N = 900\), substitute into the differential equation to find \(k\):

\(60 = k(900 - 150)\)

\(60 = 750k\)

\(k = 0.08\)

Separate variables and integrate:

\(\int \frac{1}{N - 150} \, dN = \int 0.08 \, dt\)

\(\ln(N - 150) = 0.08t + c\)

When \(t = 0\), \(N = 650\):

\(\ln(650 - 150) = c\)

\(\ln 500 = c\)

Thus, \(\ln(N - 150) = 0.08t + \ln 500\)

\(N - 150 = 500e^{0.08t}\)

\(N = 500e^{0.08t} + 150\)

(iii) Substitute \(t = 15\) into the expression for \(N\):

\(N = 500e^{0.08 \times 15} + 150\)

\(N \approx 1810\)

Alternatively, solve for \(t\) when \(N = 2000\):

\(2000 = 500e^{0.08t} + 150\)

\(1850 = 500e^{0.08t}\)

\(e^{0.08t} = 3.7\)

\(0.08t = \ln 3.7\)

\(t \approx 16.4\)

Since \(t = 16.4\) is greater than 15, the target is not met.