I definitely keep getting 8.65 not 8.95. The way I did it was treat the ammonium salt as the associated version of the acid so (NH4)2SO4, this dissociates (assuming) into 2 moles of ammonium ions, therefore the concentration of NH4+ is 0.4M. So I get Ka={[H+][NH3]/2[(NH4)2SO4]} rearranging to find pH I get -log((Ka x 2[(NH4)2SO4]/[NH3])) which gives me 8.649751982 which is roughly 8.65.
Right I see where the answer comes from, the question assumes that the salt acts to dissociate to give 1 mole of ammonium ions rather than 2. So, pH=-log{[Ka x [(NH4)2(SO4)]/[NH3]} = 8.950781977 = 8.95 2sf