If dry ammonia gas is passed through anhydrous copper sulfate, will it turn blue (due to the formation of tetraamminecopper(II) complex)? Or will silver chloride form diamminesilver(I) complex in liquid ammonia?
If the above complexes do not get formed in non-aqueous mediums, why is it so?
It should be noted that ammonia is a mildly polar solvent and that (for instance) you can prepare $\ce{[Cu(NH3)6]^2+}$ by reaction with liquid ammonia1. The description of the geometry (two pages earlier) of $\ce{CuSO4.5H2O}$ (namely containing a square planar $\ce{[Cu(H2O)4]^2+}$ unit with two additional $\ce{[SO4]^2-}$ above and below the plane) suggests that ammine ligands could fulfill a similar role to water.
Bear in mind that passing ammonia gas over a solid will probably mean that if the reaction happens it will only happen near grain surfaces.
To answer your second suggestion, Housecroft and Sharpe1 identifies a reaction producing $\ce{[Ag(NH3)4]+}$ from $\ce{Ag2O}$ in liquid $\ce{NH3}$.
Hope this is informative.
[1] Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 2nd Edition; Pearson Prentice Hall, 2005; pp. 635–637, 693.
I cannot answer concerning the $\ce{[Ag(NH3)2]+}$ complex, because I don’t know enough about it.
I do know, however, that the tetraamminecopper(II) complex is better considered a tetraamminediaquacopper(II) complex $\ce{[Cu(H2O)2(NH3)4]^2+}$. Structurally, the four ammine ligands form an almost square planar geometry around the copper cation and have rather short bond lengths. The two aqua ligands are on the axial positions (trans if you wish) and have a significantly larger bond length due to Jahn-Teller distortion. The complex is in-between octahedral and square planar geometry. Professor Klüfers’ web scriptum to his general and inorganic chemistry course gives $\ce{Cu-N}$ bond lengths of $203~\mathrm{pm}$ and $\ce{Cu-O}$ bond lengths of $251~\mathrm{pm}$. This is consistant with $\ce{[Cu(H2O)6](ClO4)2}$ where the equatorial oxygens are $195~\mathrm{pm}$ from the central copper ion and the axial ones are $238~\mathrm{pm}$.
In liquid ammonia, there are no water molecules present that could be used to generate this complex. Instead, as Richard correctly stated, a hexaamminecopper(II) complex is formed. You could also argue that the concentration of ammonia is high enough in liquid ammonia (compare to ammonia concentrations in aquaeous solutions) that all six coordination sites can be filled.
Generally, there is no restriction for coordination compounds to form only in aquaeous solutions. Well-known examples would include every single palladium-catalysed organic reaction in organic sovlents, the Sharpless epoxidation which uses a $\ce{[Ti(tartrate)(OR)2]2}$ catalyst and many organometallic compounds that exist in coordination clusters in solution. In one lab course, I was assigned to generate $\ce{[Ni(DMSO)6][NiCl4]}$ (termed $\ce{NiCl2 . 3 DMSO}$) from anhydrous $\ce{NiCl2}$ in DMSO by dissolving and filtration.
