Answer:
I and IV
Step-by-step explanation:
Increasing the number of particles at one side of the reaction (H2 in this case) results in the shifting of the equilibrium to the side with lesser number of particles, so in this case the equilibrium will shift to the left (towards the reactants)
A decrease in temperature will always function to favor the exothermic reaction, and since the backwards reaction is exothermic, the equilibrium will shift to the left (towards the reactants).
Option II and V will shift the equilibrium to the products, and adding a catalyst has no effect on the equilibrium.
Hope this helps!