1. This one is tricky, because there are two possible outcomes, depending on the concentration of the ammonia:
a) If the concentration of NH3 is not too high, you'll form a precipitate of Cu(OH)2(s). Here are the reactions:
Hydrolysis reactions of the weak base NH3 (an equilibrium):
NH3(aq) + H2O(l) 
NH4+(aq) + OH-(aq)
Reaction of Cu2+ ion from CuSO4(aq) with the hydroxide ions:
CuSO4(aq) + 2OH-(aq) --> Cu(OH)2(s) + SO42-(aq)
Or overall:
CuSO4(aq) + 2NH3(aq) + 2H2O(l) --> Cu(OH)2(s) + (NH4)2SO4(aq)
b) If the concentration of NH3 is higher, the soluble, deep blue tetraamminecopper(II) complex forms:
CuSO4(aq) + 4NH3(aq) --> [Cu(NH3)4]SO4(aq)
2. >> Nickel(II) hydroxide is strongly heated... <<
This will be a simple decomposition reaction (you have it right except O2 is not needed):
heat
Ni(OH)2(s) -----> NiO(s) + H2O(g)
3. This may be in your textbook in a chapter titled something like, "Chemistry of Nonmetallic Elements" or "Descriptive Chemistry of Nonmetallic Elements". You usually get a comparison of the reaction of H2SO4 and NaF or CaF2 (forms HF), NaCl (forms HCl), NaBr (forms Br2), and NaI (forms I2). These last reactions are more complicated. With NaCl, we have
H2SO4(l) + NaCl(s) --> HCl(g) + NaHSO4(s)
4. First let's compare electron configurations before and after removing the outer six electrons:
Before: P 1s22s22p63s23p3 S 1s22s22p63s23p4
After: P6+ 1s22s22p5 
S6+ 1s22s22p6 
Removal of six electrons from sulfur gives a noble gas configuration (neon), but removal of the sixth electron from phosphorus requires removing the electron from this same noble gas configuration, requiring much more energy to do. (I6 for S is 8490 kJ/mol, while I6 for P is 21200 kJ/mol.)
Keep at it!
Steve
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