Answer: tangent \(y=-6x+14\); the other point is \((4,-10)\).

(a) When \(x=1\),

\(y=1^3-6(1)^2+3(1)+10=8.\)

Differentiate:

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

At \(x=1\),

\(\frac{dy}{dx}=3-12+3=-6.\)

So the tangent through \((1,8)\) is

\(y-8=-6(x-1).\)

Hence

\(y=-6x+14.\)

(b) At another intersection with the curve,

\(x^3-6x^2+3x+10=-6x+14.\)

So

\(x^3-6x^2+9x-4=0.\)

Since the tangent already touches the curve at \(x=1\), \(x=1\) is a repeated root. Factorising,

\(x^3-6x^2+9x-4=(x-1)^2(x-4).\)

The other intersection occurs when

\(x=4.\)

Substitute into the tangent:

\(y=-6(4)+14=-10.\)

Therefore the point where the tangent meets the curve again is

\((4,-10).\)