For the block to remain on the turntable, the force of static friction must provide the necessary centripetal force for its uniform circular motion. Let $m$ be the mass of the block, $r$ its distance from the center, $\omega$ the angular velocity, and $\mu_s$ the coefficient of static friction.

First, we analyze the forces acting on the block. In the vertical direction, the normal force $N$ balances the gravitational force $mg$:

$$N - mg = 0 \implies N = mg$$

In the horizontal (radial) direction, the static friction $f_s$ provides the centripetal force $F_c = m a_c = m \omega^2 r$:

$$f_s = m \omega^2 r$$

The block will not slide as long as the required centripetal force does not exceed the maximum static friction, $f_{s,max} = \mu_s N$.

$$f_s \le f_{s,max}$$ $$m \omega^2 r \le \mu_s N$$

Substituting $N = mg$:

$$m \omega^2 r \le \mu_s mg$$

The mass $m$ cancels out, leaving the condition on the angular velocity:

$$\omega^2 \le \frac{\mu_s g}{r}$$

[Q1] The maximum angular velocity, $\omega_{max}$, occurs when the required centripetal force is equal to the maximum available static friction.

$$\omega_{max}^2 = \frac{\mu_s g}{r}$$

Solving for $\omega_{max}$:

$$\omega_{max} = \sqrt{\frac{\mu_s g}{r}}$$

Substituting the given values ($r = 0.3$ m, $\mu_s = 0.4$, $g = 9.8$ m/s²):

$$\omega_{max} = \sqrt{\frac{0.4 \times 9.8 \, \text{m/s}^2}{0.3 \, \text{m}}} \approx 3.615 \text{ rad/s}$$

[Q2] The period of rotation $T$ is related to the angular velocity $\omega$ by $T = 2\pi / \omega$. The minimum period $T_{min}$ corresponds to the maximum angular velocity $\omega_{max}$.

$$T_{min} = \frac{2\pi}{\omega_{max}}$$

Substituting the derived expression for $\omega_{max}$:

$$T_{min} = 2\pi \sqrt{\frac{r}{\mu_s g}}$$

Substituting the calculated value of $\omega_{max}$:

$$T_{min} = \frac{2\pi}{3.615 \, \text{rad/s}} \approx 1.738 \text{ s}$$