since dilut OH- & Aq NH3 produce similar results, why does it differ when there is excess OH_ and strong NH3.

Answer 1

They produce similar results due to the presence of OH- in both compounds in aqueous solution.
NH3 reacts water according to:
NH3 + H2O <-> NH4^+ + OH^-, hence the presence of OH- in aqueous ammonia.

The difference in excess OH- and strong/concentrated NH3 is due to certain ions that can form soluble complexes only with NH3, with NH3 being the ligand, and not OH-, and vice versa.

Look at the example of Copper (II) [Cu^2+] ion, which in aqueous form gives a blue solution as in CuSO4
On addition of OH-, a blue precipitate of Cu(OH)2 is formed, insoluble in excess OH-, as this ion cannot form a soluble complex with OH-.
However, on addition of NH3, a blue precipitate of Cu(OH)2 is formed with a little NH3, but soluble in excess NH3 to give [Cu(NH3)4]^2+, which is a dark blue solution.

On the contrary, let's look at Pb2+ ions.
On adding OH-, a white precipitate of Pb(OH)2 is formed due to the presence of OH-, but the white precipitate dissolves on excess OH- due to the formation of a soluble [Pb(OH)4]^2+.
For NH3, there is a white precipitate (Pb(OH)2) due to the OH- from dissociation of aqueous NH3, but the precipitate cannot dissolve in excess NH3 due to the inability of Pb2+ ions to form a complex with NH3.

Answer 2

As I answered in your other question, OH- and NH3 can act as ligands, forming complex ions with various metals, resulting in ionic species that will be soluble in water.

For example, ferric ion will precipitate from either weak base or weak ammonia solutions to form Fe(OH)3 (ammonia reacts with water to produce ammonium and hydroxide). In excess hydroxide, it picks up three more OH-'s to form Fe(OH)6 (3-). In excess ammonia, it forms Fe(NH3)6 (3+).