Solution

$a)$ For a body to start sliding, the applied force must exceed the friction force. When determining the magnitude of the friction force, it is necessary to take into account that, in accordance with Newton's third law, the body acts on the board, and the board on the body, so a double friction force will be exerted $$F_t = \mu g (m_1 + m_2)$$ Condition for starting movement: $$\boxed{F > \mu g (m_1 + m_2)}$$ To determine the sliding time, we use kinematic equations: $$l = \frac{a t^2}{2} \Rightarrow t = \sqrt{\frac{2L}{a}}$$ The acceleration a when a constant force $F_0$ acts along the board is determined from the equation of Newton's second law in projection onto the direction of motion: $$F_0 - F_t = m_1 a$$ $$a = \frac{F_0 - \mu g (m_1 + m_2)}{m_1}$$ Sliding time: $$\boxed{t = \sqrt{\frac{2 L m_1}{F_0 - \mu g (m_1 + m_2)}}}$$

$b)$ The body is acted upon by two external forces $F_0$ and the friction force $F_{tr}$. Let us write down Newton's second law for the body $$m_1a_1 = F_0 - F_{тр}$$ Considering that $$F_{тр} = \mu N = \mu m_1 g$$ $$m_1a_1 = F_0 - \mu m_1 g$$ Where $a_1$ we find as $$\boxed{a_1 = \frac{F_0 - \mu m_1 g}{m_1}}$$ Meanwhile, of the external forces, only the friction force with the block acts on the board

Then Newton's second law for the board $$m_2a_2 = \mu m_1 g$$ We express $a_2$ $$\boxed{a_{2}=\mu g\frac{m_{1}}{m_{2}}}$$