Answer: \(\boxed{d=5.5}\), the speed at \(OP=7\) is \(\boxed{4\ \mathrm{m\ s^{-1}}}\), and the time from \(A\) to \(B\) is \(\boxed{0.258\ \mathrm{s}}\).

(i) In SHM, the magnitude of the acceleration is \(\omega^2\) times the distance from the centre of motion.

The centre is \(d\) metres from \(O\). Therefore the distances of \(A\) and \(B\) from the centre are

\[6.5-d\quad\text{and}\quad 7.5-d.\]

The acceleration at \(B\) is twice the acceleration at \(A\), so

\[\omega^2(7.5-d)=2\omega^2(6.5-d).\]

Cancel \(\omega^2\):

\[7.5-d=13-2d.\]

Hence

\[\boxed{d=5.5}.\]

(ii) The period is \(\pi\) seconds, so

\[\omega=\frac{2\pi}{\pi}=2.\]

The maximum acceleration in SHM is \(\omega^2a\), where \(a\) is the amplitude. Therefore

\[10=2^2a,\qquad a=2.5.\]

When \(OP=7\), the displacement from the centre is

\[x=7-5.5=1.5.\]

Use the SHM speed formula

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

So

\[v^2=2^2(2.5^2-1.5^2)=4(6.25-2.25)=16.\]

Thus

\[v=\boxed{4\ \mathrm{m\ s^{-1}}}.\]

(iii) Relative to the centre, the positions of \(A\) and \(B\) are

\[x_A=6.5-5.5=1,\qquad x_B=7.5-5.5=2.\]

Use \(x=a\sin\omega t\). The direct time from \(A\) to \(B\) is

\[t=\frac{1}{\omega}\left(\sin^{-1}\frac{x_B}{a}-\sin^{-1}\frac{x_A}{a}\right).\]

Substitute \(a=2.5\) and \(\omega=2\):

\[t=\frac12\left(\sin^{-1}0.8-\sin^{-1}0.4\right).\]

So

\[t=\frac12(0.9273-0.4115)=0.2579\ldots.\]

Therefore the time is \(\boxed{0.258\ \mathrm{s}}\) to 3 significant figures.