Le chatelier's principle help

The problem:
N2(g) + 3H2(g) <-> 2NH3(g) + 22.0 kcal

What happens to the concentrations (choices are: increase/decrease/same) N2 / H2 / NH3 when:
1. The temperature is increased.
2. The pressure is increased.
3. NH3 is removed.
4. H2 is removed.

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Here are my questions / what I don't quite get:
1. For number 1; My notes say that the reaction shifts away from the heat, so does that mean that BOTH N2 and H2 increase?

2. For number 2; My notes say the reaction will shift toward the side with the least gaseous moles (that would be the product side), so does that mean that NH3 increases, and N2 / H2 stays the same?

3. For number 3; My notes say the reaction will shift away from what is removed, so does NH3 stay the same and both N2 and H2 increase?

4. For number 4; Does N2 stay the same and NH3 increase?

My teacher wasn't clear at all about this - we went over what happens to the overall shift but not each individual concentration of reactants/products. Thank you in advanced.

A + B <-> C + D

Left side: reactants / forward reaction
Right side: products / reverse reaction

1. According to Le Chatelier's principle, the system should counteract an increase in temp. by decreasing the temp.

This is an exothermic reaction. In exothermic reactions, an increase in temperature will favour the reactants. This is because the products are heat and NH3, and the system gets rid of heat to try and restore the system to what it was before the increase. So yes, both N2 and H2 increase (whatever is on the reactants side)

However, in endothermic reactions, then increase in temperature favours the products.

2. No; if NH3 increases, N2/ H2 cannot stay the same. This is because you're assuming NH3 is produced out of nothing. The reaction states that N2 + 3H2 is used to make 2NH3. So N2 and 3H2 are used up to make the 2NH3. Remember, Le Chatelier's principle favours reactions (what actually happens). In this case, it favours the forward reaction (N2 + 3H2 -> 2NH3)