If $x=5\sin \left(\pi t+\frac{\pi }{3}\right)\text{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion respectively, are:

The standard equation for SHM is:
$$x=A\sin (\omega t+\varphi )$$
Comparing the given equation $x=5\sin \left(\pi t+\frac{\pi }{3}\right)$ with the standard one:Amplitude ($A$): $A=5\text{m}$ Angular Frequency ($\omega$): $\omega =\pi \text{rad/s}$ We know that $\omega =\frac{2\pi }{T}$, so:
$$T=\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2\text{s}$$
Thus, Amplitude $=5\text{m}$ and Time Period $=2\text{s}$.