Answer: \(y=x^3-2x^2-4x+8\); stationary points \(\left(-\frac23,\frac{256}{27}\right)\) and \((2,0)\); \(0\lt k\lt \frac{256}{27}\).

(a) Expand the expression:

\((x^2-4)(x-2)=x^3-2x^2-4x+8.\)

So

\(a=-2,\qquad b=-4.\)

Differentiate:

\(\frac{dy}{dx}=3x^2-4x-4.\)

At stationary points,

\(3x^2-4x-4=0.\)

Factorise:

\((3x+2)(x-2)=0.\)

Thus

\(x=-\frac23 \quad\text{or}\quad x=2.\)

Substituting into the curve gives

and

\(y=0 \quad\text{when}\quad x=2.\)

So the stationary points are

(b) The intercepts of

\(y=\left|(x^2-4)(x-2)\right|\)

with the \(x\)-axis occur at

\(x=-2,\qquad x=2.\)

The \(y\)-intercept is

\(y=|(0^2-4)(0-2)|=8.\)

The graph is never below the \(x\)-axis. It has a sharp point at \((-2,0)\), a local maximum at \(\left(-\frac23,\frac{256}{27}\right)\), and touches the \(x\)-axis at \((2,0)\).

(c) For the horizontal line \(y=k\) to meet the graph in exactly four different points, it must lie above the \(x\)-axis but below the local maximum.

Therefore

\(0\lt k\lt \frac{256}{27}.\)