H₂SO₄ is an acid, so it will add H⁺ to the solution.  NaOH is a base, so it will remove H⁺ from the solution.  This should point you toward the answer.

I am trying to figure out the direction of shift in equilibrium when adding H2SO4 to the equation 2CrO4 + 2H = Cr2O7 + H2O.  After that, the shift by adding NaOH.  I understand why the shift occurs and what happens when it does, but am confused as to the direction of shift.  

you must remember that the adaption will be such that equilibrium maintains --> K doesn't change.

so if we add H2SO4 and thus H+ we have more H+ on the left. There are more effective collisions between chromate and H+ resulting in the equilibrium shift to the right.

Another method:

K = [Cr2O7(2-)] / [CrO4(-)][H+]^2

when we add H+ the denominator of K rises, resulting in a lowering of the equilibriumconstant. The equilibrium will shift to that side such that the change will be undone and there is no effect on the system. To accomplish this, the numerator must rise and thus the equilibrium will shift to the right (the side of the numerator)--> [dichromate ion] rises.

When we add a base, we capture H(+) away from the system, resulting in less effective collisions between chromate and H+, resulting in the equilibrium shifting to the left.

Or: [H+] decreases, resulting in a rise of the equilibriumconstant. Therefore the numerator must decrease, so the equilibrium will shift to the left.

heating the system will let all reaction go faster, since rising the temperature will let the molecules move faster and thus more effective collisions. But the endothermic reaction in particular will run 'more faster then faster'.