For general advice, I would try to make sure you understand a concept before you just start entering things into equations without knowing why you're doing it. If you're just entering things into you're calculator without really knowing what you're doing, you'll have more trouble doing the problems, and newer concepts will be harder to learn, because you won't be able to apply your previous knowledge to new ideas. Try talking to you're teacher if you're not getting something, or if your teacher isn't very good at their job, look it up online (note, Wikipedia is not a very good resource for this, as science articles tend to not be written in a way so that they're easily understandable by people outside of that field).
Also, if you're having trouble, do more problems. There's no reason to limit yourself to just the problems you're assigned if you're struggling. The more problems you do, the better you are going to learn the material, period.
Anyway, more specifically, I think I might be able to help you with pH and pOH of weak acids and bases. I'm not sure exactly how much you know about this, so I'm just going to start from the beginning.
Strong acids and bases dissociate (seperate, basically) completely in water. For example, HCl will completely dissociate into H+ (well, H3O+ really, but it doesn't make much of a difference) and Cl- ions. Therefore, the concentration of H+ ions is going to be the same as the initial concentration of HCl, which you can then use to calculate the pH.
Weak acids and bases, on the other hand, only partially dissociate in water. For example, in water, some acetic acid (CH3CO2H) dissociates into CH3CO2- and H+ ions, and some of it doesn't. Basically, a weak acid dissolving in water is part of a reversible reaction, where as the reactants turn into products (the forward reaction), some of the products turn back into reactants (the reverse reaction). At a certain point, the rate of the forward and reverse reactions is the same, and the amount of products and reactants doesn't change. At this point, the reaction is said to be in equilibrium.
Now, the ratio of the products and reactants in a reversible reaction is always the same, regardless of how much of them you start with. This ration is the equilibrium constant, which for acids is K(a), and K(b) for bases (note, it should be a subscript instead of in parentheses). The equation for K(a) or K(b) is:
K(a) = (concentrations of products) / (concentration of reactants)
Now, for acetic acid, the chemical equation is:
CH3CO2H -> CH3CO2- + H+
For the calculation of K(a), the concentration of each part in either the products or reactants should be multiplied together. So, for acetic acid:
K(a) = ((Concentration of CH3CO2-) * (Concentration of H+)) / (Concentration of CH3CO2H)
So, everything except the concentration of H+ is known:
Concentration of H+ = (K(a) * (Concentration of CH3CO2H)) / (Concentration of CH3CO2)
Now, I'm guessing most of your problems are along the lines of "what is the pH of a 0.1 M acetic acid solution, where K(a) = 1.8 * 10^-5." To solve this, first, you need to determine what the concentrations of each chemical in the reaction. To do this, a table can be helpful. Because the change in concentration for each chemical is unknown, but each change in concentration is equal, we can use x to represent the change in concentration
CH3CO2H H+ CH3CO2-
Initial concentration 0.1 0 0
Final concentration 0.1 - x x x
Then, substitute the values in the table into the K(a) equation.
1.8 * 10^-5 = (x * x) / (0.1 - x) = x^2 / (0.1 - x)
Then, we try to get the equation in the form of a quadratic equation (ax^2 + bx + c = 0)
-x^2 + (1.8 * 10^-5)x ((1.8 * 10^-5) * 0.1) = 0
Then we can use the quadratic formula to solve for x.
x = 6.8 x 10^-4
Then, because the concentration of H+ = x, we can simply substitute x into the pH equation to find the pH of the solution.
pH = -log(6.8 x 10^-4) = 3.2
I hope that helps make things a little clearer, although you still should probably talk to your teacher (again, if they're any good at teaching).
