A particle in simple harmonic motion has a period of 8.0 s and an amplitude of 5.0 m. Calculate its speed when it is 3.0 m from the center and its maximum speed.

The angular frequency is related to the period by the formula:

$$\omega = \frac{2\pi}{T}$$

Substitute the given period:

$$\omega = \frac{2\pi}{8.0 \text{ s}} = \frac{\pi}{4.0} \text{ rad/s}$$ $$\omega \approx 0.7854 \text{ rad/s}$$

The speed ($v$) of a particle in SHM at a displacement ($x$) from the center is given by:

$$v = \omega \sqrt{A^2 - x^2}$$

Given $x = 3.0$ m and $A = 5.0$ m.

$$v = \frac{\pi}{4.0} \sqrt{(5.0 \text{ m})^2 - (3.0 \text{ m})^2}$$ $$v = \frac{\pi}{4.0} \sqrt{25 \text{ m}^2 - 9 \text{ m}^2}$$ $$v = \frac{\pi}{4.0} \sqrt{16 \text{ m}^2}$$ $$v = \frac{\pi}{4.0} \times 4.0 \text{ m/s}$$ $$v = \pi \text{ m/s}$$ $$v \approx 3.14 \text{ m/s}$$

The speed of the particle when it is 3.0 m from the center is $\boxed{\text{3.14 m/s}}$.

The maximum speed ($v_{max}$) occurs at the equilibrium position ($x=0$) and is given by:

$$v_{max} = \omega A$$

Substitute the values for $\omega$ and $A$:

$$v_{max} = \left(\frac{\pi}{4.0} \text{ rad/s}\right) \times (5.0 \text{ m})$$ $$v_{max} = \frac{5\pi}{4.0} \text{ m/s}$$ $$v_{max} \approx 3.93 \text{ m/s}$$

The maximum speed is $\boxed{\text{3.93 m/s}}$.

The maximum acceleration ($a_{max}$) occurs at the extreme positions ($x=\pm A$) and is given by:

$$a_{max} = \omega^2 A$$

Substitute the values for $\omega$ and $A$:

$$a_{max} = \left(\frac{\pi}{4.0} \text{ rad/s}\right)^2 \times (5.0 \text{ m})$$ $$a_{max} = \frac{\pi^2}{16} \times 5.0 \text{ m/s}^2$$ $$a_{max} = \frac{5\pi^2}{16} \text{ m/s}^2$$ $$a_{max} \approx 3.08 \text{ m/s}^2$$

The maximum acceleration is $\boxed{\text{3.08 m/s}^2}$.