Answer: The speed at \(O\) is \(\boxed{2\ \mathrm{m\ s^{-1}}}\), and the time from \(A\) to \(B\) is \(\boxed{2.64\ \mathrm{s}}\).

Let the amplitude be \(R\), and let \(\omega\) be the angular frequency.

In SHM, the speed at distance \(x\) from the centre is

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

At \(A\), the distance from \(O\) is \(3.5\), so

\[v_A^2=\omega^2(R^2-3.5^2).\]

At \(B\), the distance from \(O\) is \(1\), so

\[v_B^2=\omega^2(R^2-1^2).\]

We are told that \(v_B=2v_A\). Hence

\[\omega^2(R^2-1)=4\omega^2(R^2-3.5^2).\]

Cancel \(\omega^2\):

\[R^2-1=4(R^2-12.25).\]

So

\[R^2-1=4R^2-49,\]

and therefore

\[3R^2=48,\qquad R=4.\]

The maximum acceleration in SHM is \(\omega^2R\). This is given as \(1\), so

\[\omega^2(4)=1,\qquad \omega=\frac12.\]

The speed at the centre \(O\) is the maximum speed:

\[v_O=\omega R=\frac12(4)=2.\]

For the time from \(A\) to \(B\), use \(x=R\sin\phi\). Moving directly from \(A\) to \(B\), the phase change is

\[\sin^{-1}\left(\frac{3.5}{4}\right)+\sin^{-1}\left(\frac{1}{4}\right).\]

Thus

\[\omega t=\sin^{-1}(0.875)+\sin^{-1}(0.25)=1.065+0.253=1.318.\]

Since \(\omega=\frac12\),

\[t=\frac{1.318}{1/2}=2.636\ldots.\]

So the time taken is \(\boxed{2.64\ \mathrm{s}}\).