To understand how the changes in pressure affect the position of the equilibrium stick to this example and imagine you increase the pressure of the reaction vessel. It should be apparent that the right side of the equation will be more 'uncomfortable' with that change as there is more gas on that side. These two moles of carbon monoxide will become more 'squeezed' with the increased pressure and the equilibrium will move to the left.
Let's state clearly according to Le Chatelier:
If you increase the pressure, the equilibrium will shift to the side with
least moles of gas.
If you decrease the pressure, the equilibrium will shift to the side with
most moles of gas.
This should help with points 2 and 4.
1, 3, 5 you have absolutely correct!