Answer: (a) \(y=-x+2\); (b) \(x=3-\sqrt2\) or \(x=3+\sqrt2\).

We start with the main method. Differentiate to obtain the tangent gradient, then use the negative reciprocal for the normal gradient where needed.

(a) The curve is

\(y=x^3-7x^2+12x-5.\)

Differentiate:

\(\frac{\mathrm dy}{\mathrm dx}=3x^2-14x+12.\)

At \(x=1\),

\(\frac{\mathrm dy}{\mathrm dx}=3-14+12=1.\)

So the tangent gradient is \(1\), and the normal gradient is \(-1\).

The normal passes through \((1,1)\), so

\(y-1=-1(x-1).\)

Hence

\(y=-x+2.\)

(b) The other intersections of the normal and the curve satisfy

\(x^3-7x^2+12x-5=-x+2.\)

Rearranging gives

\(x^3-7x^2+13x-7=0.\)

Since the normal touches the curve at \(x=1\), \(x=1\) is one root. Divide by \(x-1\):

\(x^3-7x^2+13x-7=(x-1)(x^2-6x+7).\)

The other intersections therefore come from

\(x^2-6x+7=0.\)

Using the quadratic formula,

\(x=\frac{6\pm\sqrt{36-28}}2.\)

So

\(x=\frac{6\pm\sqrt8}{2}=3\pm\sqrt2.\)

Therefore the required \(x\)-coordinates are

\(x=3-\sqrt2\quad\text{and}\quad x=3+\sqrt2.\)

This completes the solution and gives the required result.