Answer: \(\boxed{OM=2\ \mathrm{m}}\), \(\boxed{T=8\ \mathrm{s}}\), and \(\boxed{v_L=\dfrac{\pi}{2}\ \mathrm{m\ s^{-1}}}\).

(i) In simple harmonic motion, the magnitude of the acceleration is

\[\omega^2|x|.\]

So the ratio of the acceleration magnitudes at \(L\) and \(M\) is the same as the ratio of their distances from \(O\):

\[\frac{OL}{OM}=\frac34.\]

Since \(OL=1.5\),

\[\frac{1.5}{OM}=\frac34.\]

Thus

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

(ii) The amplitude is \(2.5\) m. Write the motion in the form

\[x=2.5\sin(\omega t).\]

To travel directly from \(L\) to \(M\), the particle passes from one side of \(O\) to the other, so the total phase angle covered is

\[\sin^{-1}\left(\frac{1.5}{2.5}\right)+\sin^{-1}\left(\frac{2}{2.5}\right).\]

The time taken is 2 s, so

\[2\omega=\sin^{-1}(0.6)+\sin^{-1}(0.8).\]

These two angles add to \(\frac{\pi}{2}\), so

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

Hence

\[\omega=\frac{\pi}{4}.\]

The period is

\[T=\frac{2\pi}{\omega}=\frac{2\pi}{\pi/4}=8.\]

Therefore the period is \(\boxed{8\ \mathrm{s}}\).

(iii) The speed in SHM at distance \(x\) from the centre is

\[v=\omega\sqrt{a^2-x^2}.\]

At \(L\), \(x=1.5\) and the amplitude is \(a=2.5\). Therefore

\[v_L=\frac{\pi}{4}\sqrt{2.5^2-1.5^2}.\]

So

\[v_L=\frac{\pi}{4}\sqrt{6.25-2.25}=\frac{\pi}{4}\cdot2=\boxed{\frac{\pi}{2}\ \mathrm{m\ s^{-1}}}.\]