I have tried and tried to use the henderson-hasselbalch equation for this problem, but i do not understand how to plug in the data: how much ammonium sulfate should be added to 500mL of .3M ammonia to produce a buffer with a pH of 9.?

You need to confirm all of this; your text may use slightly different Ka, Kb, molar masses, etc.

I use 1.8E-5 for Kb for NH3; therefore, I calcualte 9.26 for Ka.
pH = pKa + log [(base)/(acid)]
9.00 = 9.26 + log (b/a)
b/a = 0.55 or
base = 0.55*acid
500 mL x 0.3M = 150 mmoles NH3.
acid = 150 mmols/0.55 = 273 mmoles acid or 273/2 = 136 mmoles (NH4)2SO4 (since there are two moles NH4^+/mole of the ammonium sulfate.
moles = g/molar mass or
g = moles x molar mass = 0.135 x 132 = about 18 g.

I like to check these things to make sure the pH is what I want.
18 g (NH4)2SO4 = 18/132 = 0.135 moles (NH4)2SO4 = 0.272 moles NH4^+

pH = 9.26 + log(0.150/0.272)
pH = 9.26 + (-0.26) = 9.00 :-)
I hope this helps.