Edit: Here are all the problems and my new answers to them so you can see what work I've done. A picture of the problem can be found on page 26, in question 13 here, though the problem is different.
Here are all the problems. I'll post them with my new attempts and then maybe someone can tell me if and where I am going wrong.
(a) Assume that the coefficient of static friction between the board and the box is not
known at this point. What is the magnitude, and the direction, of the acceleration of
the box in terms of $F_{\text{f}}$?
If $a_{\text{box}}$ is the acceleration on the box, then
$F_{\text{f}}=m_{1}a_{\text{box}}$ so $a_{\text{box}}=\frac{F_{\text{f}}}{m_{1}}$
This must act in the direction of $F$ or the box would slide or accelerate off the negative side of the board (taking the direction of $F$ to be positive)
(b) Now take the coefficient of static friction to be $\mu_{S}$. What is the largest possible magnitude of acceleration of the box? Express your answer in terms of some or all of the variables $F, \mu_{S}, m_{1}, m_{2}$ and $ g$.
So $ m_{1} a_{\text{box}} = F_{\text{f}} \leq \mu_{S} F_{n}$ where $ F_{n} $ is the normal force on the box. $ F_{n} = m_{1}g $ so $ m_{1} a_{\text{box}} \leq \mu_{S} m_{1}g$
$a_{\text{box}} \leq \mu_{S}g$ and therefore $\max{a_{\text{box}}} = \mu_{S}g$
(c) Write down the sum of all horizontal forces on the board, taking the positive direction
to be towards the right. Give your answer in terms of $F$ and $F_{f}$.
The total force on the board, $F_{\text{board}}$, is equal to the accelerating force, $F$, plus the friction from the box acting on the board, $-F_{f}$ i.e. $F_{\text{board}}=F-F_{f}$
(d) Find an expression for the acceleration of the board when the force of static friction reaches its maximum possible value in terms of $F, \mu_{S}, m_{1}, m_{2}$ and $ g$.
$\max{F_{\text{f}}}=\mu_{S}m_{1}g$ so the total force on the board when static friction is at its maximum is $F_{\text{board}}=F-\mu_{S}m_{1}g$. Because the mass of the box $m_{1}$ acts through the board, $F_{\text{board}}=(m_{1}+m_{2})a_{\text{board}}$, where $a_{\text{board}}$ is the acceleration of the board.
This means $(m_{1}+m_{2})a_{board} = F-\mu_{S}m_{1}g$ so $a_{board} = \frac{F-\mu_{S}m_{1}g}{m_{1}+m_{2}}$
What is the minimum value of the constant force, $F$, applied to the board, needed to ensure that the accelerations satisfy $|a_{\text{board}}|>|a_{\text{box}}|$? Express your answer in terms of some or all of the variables $\mu_{S}, m_{1}, m_{2}, g$ and $L$. Do not include $F_{\text{f}}$ in your answer.
Assuming that the force of static friction is at its maximum when slippage occurs, $\left|\frac{F-\mu_{S}m_{1}g}{m_{1}+m_{2}}\right|> |\mu_{S}g|$
If $F<\mu_{S}m_{1}g$ then $\mu_{S}m_{1}g - F > \mu_{S}m_{1}g + \mu_{S}m_{2}g $
then $-F > \mu_{S}m_{2}g$ and $ F < -\mu_{S}m_{1}g$. This implies F is negative when it is actually positive so $F \geq \mu_{S}m_{1}g$ and then
$F-\mu_{S}m_{1}g > \mu_{S}m_{1}g + \mu_{S}m_{2}g $
and $F>\mu_{S}g(2m_{1}+m_{2})$
