[Q1] Compare the magnitude of $p_1$ and $p_2$

The pressure exerted by a container on the table is given by $p = \frac{W_{total}}{A_{base}}$, where $W_{total}$ is the total weight of the system and $A_{base}$ is the area of the container's base. Since the two containers A and B are identical, their base areas are equal ($A_1 = A_2$). Therefore, the comparison of pressures reduces to a comparison of the total weights of the two systems.

The total weight of system A (container, water, wooden sphere) is:

$$W_1 = (m_c + m_w + m_{wood})g$$

The total weight of system B (container, alcohol, copper sphere) is:

$$W_2 = (m_c + m_{alc} + m_{cu})g$$

We are given that the masses of the container ($m_c$) are identical and the masses of the liquids are equal ($m_w = m_{alc}$). The comparison of weights thus depends on the masses of the spheres, $m_{wood}$ and $m_{cu}$. The masses are related to density ($\rho$) and volume ($V$) by $m = \rho V$.

$$m_{wood} = \rho_{wood}V_{wood}$$ $$m_{cu} = \rho_{cu}V_{cu}$$

Given that the volumes are equal ($V_{wood} = V_{cu} = V_{sphere}$) and the densities are related by $\rho_{wood} < \rho_{cu}$, we can conclude that $m_{wood} < m_{cu}$.

Therefore, the total weight of system A is less than that of system B:

$$W_1 < W_2$$

Since the base areas are equal, the pressures are related as:

$$p_1 < p_2$$

[Q2] Compare the magnitude of $F_1$ and $F_2$

The buoyant force $F_1$ acts on the wooden sphere in water. According to the given density relationship, $\rho_{wood} < \rho_{water}$, which means the wooden sphere will float. For a floating object, the buoyant force is equal to the object's weight.

$$F_1 = W_{wood} = m_{wood}g = \rho_{wood}V_{sphere}g$$

The buoyant force $F_2$ acts on the copper sphere in alcohol. Since $\rho_{cu} > \rho_{alcohol}$, the copper sphere will sink and be fully submerged. The buoyant force on a submerged object is equal to the weight of the fluid it displaces (Archimedes' principle).

$$F_2 = \rho_{alcohol} g V_{submerged}$$

Since the sphere is fully submerged, the volume of displaced alcohol is equal to the volume of the sphere, $V_{submerged} = V_{sphere}$.

$$F_2 = \rho_{alcohol} g V_{sphere}$$

Now we compare $F_1$ and $F_2$:

$$F_1 = \rho_{wood} g V_{sphere}$$ $$F_2 = \rho_{alcohol} g V_{sphere}$$

We are given the density relationship $\rho_{wood} < \rho_{alcohol}$. Since $g$ and $V_{sphere}$ are the same in both expressions, it follows that:

$$F_1 < F_2$$