(a) To find the equation of the curve, integrate \(f'(x) = 1 - \frac{2}{(x-1)^3}\).

\(\int \left(1 - \frac{2}{(x-1)^3}\right) \, dx = x + \int -\frac{2}{(x-1)^3} \, dx\)

Let \(u = x-1\), then \(du = dx\). The integral becomes \(\int -\frac{2}{u^3} \, du = \frac{1}{u^2} = \frac{1}{(x-1)^2}\).

Thus, \(y = x + \frac{1}{(x-1)^2} + c\).

Given \((0, 3)\), substitute to find \(c\):

\(3 = 0 + \frac{1}{1} + c \Rightarrow c = 2\).

Therefore, \(y = x + (x-1)^2 + 2\).

(b) The gradient of the tangent at \(x = 0\) is \(f'(0) = 3\).

The equation of the tangent is \(y - 3 = 3(x - 0) \Rightarrow y = 3x + 3\).

Set \(3x + 3 = x + (x-1)^2 + 2\) to find intersection:

\(3x + 3 = x + (x-1)^2 + 2 \Rightarrow 2x + 1 = \frac{1}{(x-1)^2}\).

Multiply through by \((x-1)^2\):

\((2x + 1)(x-1)^2 = 1 \Rightarrow (2x + 1)(x-1)^2 - 1 = 0\).

(c) Substitute \(x = \frac{3}{2}\) into \((2x + 1)(x-1)^2 - 1 = 0\):

\((2 \times \frac{3}{2} + 1)(\frac{3}{2} - 1)^2 - 1 = 0\).

\((3 + 1)(\frac{1}{2})^2 - 1 = 0 \Rightarrow 4 \times \frac{1}{4} - 1 = 0\).

Thus, \(x = \frac{3}{2}\) satisfies the equation.

Find \(y\)-coordinate: \(y = \frac{3}{2} + (\frac{3}{2} - 1)^2 + 2\).

\(y = \frac{3}{2} + \frac{1}{4} + 2 = \frac{7}{2}\).