Answer: \(\boxed{k=8}\), centre \(\boxed{\dfrac{7a}{6}\text{ from }O}\), period \(\boxed{\pi\sqrt{\dfrac ag}}\), least tension \(\boxed{\dfrac23mg}\), and maximum acceleration \(\boxed{\dfrac13g}\).
(i) For the particle \(P\), the mass is \(3m\). In equilibrium, the tension in the spring balances the component of weight down the plane:
\[T=3mg\sin\alpha.\]
Since \(\sin\alpha=\frac23\),
\[T=3mg\cdot\frac23=2mg.\]
The spring has natural length \(a\) and length \(\frac54a\), so its extension is
\[\frac54a-a=\frac14a.\]
By Hooke's law,
\[T=\frac{kmg}{a}\cdot\frac14a=\frac14kmg.\]
Thus
\[\frac14kmg=2mg,\]
so
\[\boxed{k=8}.\]
(ii) Now the mass is \(2m\). Let \(y=OQ\), measured down the plane from \(O\).
The tension in the spring is
\[\frac{8mg}{a}(y-a).\]
Taking the positive direction down the plane, Newton's second law gives
\[2m\frac{d^2y}{dt^2}=2mg\sin\alpha-\frac{8mg}{a}(y-a).\]
Substitute \(\sin\alpha=\frac23\):
\[2m\frac{d^2y}{dt^2}=\frac43mg-\frac{8mg}{a}(y-a).\]
Dividing by \(2m\),
\[\frac{d^2y}{dt^2}=\frac{4g}{a}\left(\frac{7a}{6}-y\right).\]
This shows that the acceleration is directed towards \(y=\frac{7a}{6}\) and is proportional to the displacement from that point.
Let
\[x=y-\frac{7a}{6}.\]
Then
\[\frac{d^2x}{dt^2}=-\frac{4g}{a}x.\]
This is simple harmonic motion with
\[\omega^2=\frac{4g}{a}.\]
So the centre of motion is \(\boxed{\frac{7a}{6}}\) from \(O\), and the period is
\[T=\frac{2\pi}{\omega}=\frac{2\pi}{2\sqrt{g/a}}=\boxed{\pi\sqrt{\frac ag}}.\]
(iii) The particle is released from \(E\), where the spring length is \(\frac54a\). The centre of the motion is at length \(\frac76a\), so the amplitude is
\[\frac54a-\frac76a=\frac{a}{12}.\]
The least tension occurs when the spring is shortest. The shortest spring length is
\[\frac76a-\frac1{12}a=\frac{13}{12}a.\]
So the least extension is
\[\frac{13}{12}a-a=\frac{a}{12}.\]
Therefore the least tension is
\[T_{\min}=\frac{8mg}{a}\cdot\frac{a}{12}=\boxed{\frac23mg}.\]
The maximum acceleration in SHM is
\[a_{\max}=\omega^2A.\]
Here \(\omega^2=\frac{4g}{a}\) and \(A=\frac{a}{12}\), so
\[a_{\max}=\frac{4g}{a}\cdot\frac{a}{12}=\boxed{\frac13g}.\]
