Bounding the height of the toppling force on an object 
On the above question, for the upper bound I took moments about the right bottom corner, resulting in the upper bound:
$$
H_{UB} = mg\frac{b}{2F}
$$
For the lower bound, I took moments about the COG:
$$
H_{LB} = mg\frac{\mu a}{Fa-1}*
$$
Do these bounds satisfy the toppling conditions?
My thoughts are:
1) I think friction should come into account in the upper bound i.e. I should be taking moments about COG here as well.
2) However, I'm not even sure if I am allowed to take moments about the COG because I don't think it is a pivot. 
3) For the lower bound, I think I should be taking an inertial reference frame, and using the dynamic friction coefficient (reduced resistance to toppling backwards).  However, I'm finding that difficult to visualise, what force causes the toppling backwards due to movement forwards (F=ma?)

PS. These questions are set deliberately vague. You are free to make assumptions and inferences, as long at they're outlined in the solution.
*EDIT: I spotted an arithmetic error in the lower bound.  Here are my new calculations, assuming lower bound for line of action of F is below the COG:
$$
\Sigma M_{COG} = 0
$$
$$
F(\frac{a}{2}-h)-F_f(\frac{a}{2}) = 0
$$
$$
H_{LB} = \frac{mg\mu a- Fa}{2F}
$$
 A: 
1) I think friction should come into account in the upper bound i.e. I should be taking moments about COG here as well.
2) However, I'm not even sure if I am allowed to take moments about the COG because I don't think it is a pivot.

You are allowed to pick any point as a pivot and find the limiting case where the torques are balanced.  So picking the corner or the COG are possible. 
The lower corner might have an advantage that any forces through the corner (such as friction and normal) can be ignored for torques.  But if the block is sliding under the force, then it is accelerating and fictitious forces appear.  These forces can be ignored when the COG is used as a pivot.
In the problem, there are only 4 forces.  If you consider the COG for pivot, then gravity disappears, leaving 3 to consider:


*

*The applied force

*Friction

*The normal force from the table.


The first two are relatively simple.  They act horizontally at a specific distance vertically from the COG.  Only the normal is tricky.
You have assumed that it acts in the center of the base.  But the position may change.  As you apply a force on the top of the block, it will try to turn.  As it tries to turn, the block pushes harder on the table in the front than the back.  We can consider this as the point of application of the normal moving forward.  The position of the normal provides a restoring torque that prevents tipping.  This is why a wider box is more stable.   The limiting case is that the normal force acts as far forward as possible, the front corner. 
A: I believe you should use both conditions of static equilibrium: that the sum of forces vanishes and that the sum of moments of forces vanishes. 
