Reaction of Metallic Tin with NH4Cl and treatment with Sulfur
Note: this is an incomplete answer because question is unclear till now


  Metallic $\ce{Sn}$ is reacted with $\ce{NH4Cl}$..


$$\ce{2NH4Cl + Sn → SnCl2 (white ppt.) + 2NH3 (both aq and g.) + H2 ^}$$

Ammonium chloride react with tin to produce tin(II) chloride, ammonia
and hydrogen. (source)

Some tin(II) chloride will react with aqueous ammonia to form tin (II) hydroxide and amm. chloride.(source)

$$\ce{SnCl2 + 2(NH3•H2O) or NH3(aq) → Sn(OH)2 or SnO.H2O(white ppt.)  + 2NH4Cl (white ppt.)}$$


  ..and the resulting solution is treated with Sulfur.


I am suspecting that is hydrogen sulfide and not sulfur.(Are you sure?)

$$\ce{SnCl2 + H2S → SnS(dark brown ppt.) + HCl}$$

Tin(II) chloride react with hydrogen sulfide to produce tin(II) sulfide and hydrogen chloride. Tin(II) chloride - concentrated solution. Hydrogen sulfide - saturated solution. (source)

So, answer to your questions:


What is the gaseous product? 



  Hydrogen, Some ammonia gas.



What is the color of final precipitate?



  Stannous sulfide ($\ce{SnS}$)- dark brown ppt.



The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple coloured solution. What is the composition of the solution?



  $$\ce{2[AuCl4]- + 4Sn^2+ + 2H2O -&gt; [2Au + Sn(OH)2)](purple ppt.) +
3Sn^4+ + 2H+ + 8Cl-}$$
  
  (Vogel)
  
  Note: $\ce{AuCl3}$ exist as $\ce{AuCl4-}$ ions in solution.