(a) At \(x=2\), \(3x^2-6=6>0\), so \(P\) is a minimum.

(b) Integrating once, \(dy/dx=x^3-6x+c\). Stationarity at \(x=2\) gives \(0=8-12+c\), so \(c=4\). Integrating again, \[y=\frac{x^4}{4}-3x^2+4x+d.\] Since \(y(2)=3\), \(3=4-12+8+d\), giving \(d=3\). Thus \[\boxed{y=\frac{x^4}{4}-3x^2+4x+3}.\]

(c)(i) \[\frac{dy}{dx}=x^3-6x+4=(x-2)(x^2+2x-2).\] The other roots are \(\boxed{x=-1\pm\sqrt3}\).

(c)(ii) At \(x=-1+\sqrt3\), \(y\prime\prime=6-6\sqrt3<0\), so this is a maximum. At \(x=-1-\sqrt3\), \(y\prime\prime=6+6\sqrt3>0\), so this is a minimum.