Answer: \(a=6+6\cos2t\); smallest \(t=\frac{\pi}{2}\); \(x=-\frac32\cos2t+3t^2+2t+\frac32\).

(i) Acceleration is the derivative of velocity:

\(a=\frac{dv}{dt}=\frac{d}{dt}(2+6t+3\sin2t).\)

So

\(a=0+6+3\cdot2\cos2t=6+6\cos2t.\)

(ii) Set the acceleration equal to zero:

\(6+6\cos2t=0.\)

Hence

\(\cos2t=-1.\)

The smallest positive solution is

\(2t=\pi \Rightarrow t=\frac{\pi}{2}.\)

(iii) Displacement is the integral of velocity:

\(x=\int(2+6t+3\sin2t)\,dt.\)

Integrating term by term gives

\(x=2t+3t^2-\frac32\cos2t+c.\)

When \(t=0\), the particle is passing through \(O\), so \(x=0\). Therefore

\(0=0+0-\frac32+c,\)

which gives

\(c=\frac32.\)

Hence

\(x=-\frac32\cos2t+3t^2+2t+\frac32.\)