Aluminum hydroxide reacts with an excess of hydroxide ions to form the complex ion Al(OH)4-

Question

a)Write an equation for this reaction
b) Calculate K.
c) Determine the solubility of Al(OH)3 in (mol/L) at pH 12.0

My equation was
Al(OH)3 + OH- --> Al(OH)4-
but I'm wondering if I may need to balance out the OH- with H+?
Also, I'm a little confused on what K is. I know it's an equilibrium constant, but how does it differ from Ksp or Kc or Kp? Do you have to decide which constant you're looking for? And how do I find this once I determine the right constant?

Your equation is correct for the formation of Al(OH)4^-. There is nothing to balance out; i.e., atoms balance, charge balances, so nothing needs to be added.
The formation constant is just another equilibrium constant but it pertains to the formation of complex ions; therefore, they have given the name of formation constant to it just as they gave solubility constant to the solubility of ppts. The K expression is the same as all the others; that is, (products)/(reactants) with each raised to a power indicated by the coefficient.
I don't know how to calculate K with "nothing" given.
If you find K, however, then
K = [(Al(OH)4^-)/(OH^-)]
Plug in OH^- and knowing K calculate (Al(OH)4^-.

I made a typo above.
....to it just as they gave solubility constant...... should read
....to it just as they gave solubility PRODUCT constant.......