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