Answer: The amplitude is \(\boxed{2\ \mathrm{m}}\), the period is \(\boxed{12\ \mathrm{s}}\), and the time from \(A\) to \(B\) is \(\boxed{3\ \mathrm{s}}\).

(i) In simple harmonic motion, the speed at displacement \(x\) from the centre is given by

\[v^2=\omega^2(a^2-x^2),\]

where \(a\) is the amplitude.

At \(A\), \(x=1.6\), and at \(B\), \(x=1.2\). The ratio of the speeds is

\[v_A:v_B=3:4.\]

Therefore

\[\frac{v_A^2}{v_B^2}=\frac{9}{16}.\]

Using the SHM speed formula,

\[\frac{9}{16}=\frac{a^2-1.6^2}{a^2-1.2^2}.\]

So

\[16(a^2-2.56)=9(a^2-1.44).\]

Expand and simplify:

\[16a^2-40.96=9a^2-12.96.\]

Hence

\[7a^2=28,\qquad a^2=4.\]

Therefore the amplitude is

\[\boxed{a=2\ \mathrm{m}}.\]

(ii) The maximum speed in SHM is

\[v_{\max}=a\omega.\]

We are given \(v_{\max}=\frac13\pi\), and we have found \(a=2\). Hence

\[2\omega=\frac13\pi,\qquad \omega=\frac{\pi}{6}.\]

The period is

\[T=\frac{2\pi}{\omega}=\frac{2\pi}{\pi/6}=12.\]

So

\[\boxed{T=12\ \mathrm{s}}.\]

(iii) To move directly from \(A\) to \(B\), the particle travels from one side of \(O\) to the other.

Using \(x=a\sin\omega t\), the phase angle needed is

\[\sin^{-1}\left(\frac{1.6}{2}\right)+\sin^{-1}\left(\frac{1.2}{2}\right).\]

So

\[\omega t=\sin^{-1}(0.8)+\sin^{-1}(0.6).\]

These angles are complementary, so their sum is \(\frac{\pi}{2}\). Therefore

\[t=\frac{\pi/2}{\pi/6}=3.\]

Hence the required time is

\[\boxed{3\ \mathrm{s}}.\]