[Heisenberg97]

Why is Ni(NH3)6^2+ stable and Ni(PF3)6^2+ unstable?

Attempt - have drawn d splitting of both, NH3 is weak field ligand so high spin. PF3 is strong field ligand, so low spin. Ni(NH3)6 is paramagnetic with 2 unpaired electrons, whilst Ni(PF3)6 is diamagnetic with no unpaired electrons. Parallel unpaired electrons in Ni(NH3)62+ stabilise the molecule. Really not sure where to go from here?

Why is Ni(PF3)4 stable and Ni(NH3)4 isn't?

Attempt - Again, drawn d splittings, both d10 and tetrahedral? Stuck again and based on same principle as above.

[Flatbutterfly]

Consider the HOMO-LUMO interactions namely the oxidation state of the Ni atom in the two cases versus the σ-donor –
π-acceptor properties of the NH3 and PF3 ligands.
(Even if the Ni(II)-PF3 bond were strong, [Ni(PF3)6]^2+ would probably not be stable for steric reasons.)

[Flatbutterfly]

Some more thoughts:
You could approach this problem from a Hard, Soft, Acid, Base angle (HSAB Wikipedia)
Are Ni(0), Ni^2+, NH3, and PF3 hard or soft?
Hard acids prefer hard bases and soft acids prefer soft bases.

PF3 has virtually identical bonding properties as CO.  Ni(CO)4 is a relative stable, straw-colored liquid that is exceedingly poisonous. On the other hand [Ni(CO)6]^2+ does not exist indicating that steric interactions are not responsible for the nonexistence of [Ni(PF3)6]^2+.