Answer: The amplitude is \(\boxed{2.6\ \mathrm{m}}\), and the period is \(\boxed{8\ \mathrm{s}}\).
Let \(a\) be the amplitude and let \(\omega\) be the angular speed of the SHM.
The displacement from the centre is measured from the middle of the motion. At the highest point, the displacement is \(a\). Therefore:
when the piston is \(1.6\) m below the highest point, its displacement from the centre is \(a-1.6\);
when the piston is \(0.2\) m below the highest point, its displacement from the centre is \(a-0.2\).
In SHM,
\[v^2=\omega^2(a^2-x^2).\]
Using the first given speed,
\[\left(\frac{3\pi}{5}\right)^2=\omega^2\{a^2-(a-1.6)^2\}.\]
Since
\[a^2-(a-1.6)^2=3.2a-2.56,\]
we have
\[\left(\frac{3\pi}{5}\right)^2=\omega^2(3.2a-2.56).\]
Using the second given speed,
\[\left(\frac{\pi}{4}\right)^2=\omega^2\{a^2-(a-0.2)^2\}.\]
Since
\[a^2-(a-0.2)^2=0.4a-0.04,\]
we also have
\[\left(\frac{\pi}{4}\right)^2=\omega^2(0.4a-0.04).\]
Divide the two equations to eliminate \(\omega^2\), or cross-multiply:
\[\left(\frac{3\pi}{5}\right)^2(0.4a-0.04)=\left(\frac{\pi}{4}\right)^2(3.2a-2.56).\]
Cancel \(\pi^2\):
\[\frac{9}{25}(0.4a-0.04)=\frac{1}{16}(3.2a-2.56).\]
This gives
\[0.144a-0.0144=0.2a-0.16,\]
so
\[0.1456=0.056a.\]
Hence
\[a=2.6.\]
Now substitute \(a=2.6\) into
\[\left(\frac{\pi}{4}\right)^2=\omega^2(0.4a-0.04).\]
Since
\[0.4(2.6)-0.04=1,\]
we get
\[\omega^2=\left(\frac{\pi}{4}\right)^2,\qquad \omega=\frac{\pi}{4}.\]
The period is
\[T=\frac{2\pi}{\omega}=\frac{2\pi}{\pi/4}=8.\]
Therefore the period is
\[\boxed{8\ \mathrm{s}}.\]
