Step 1: Plan the check for each complex.
A complex is diamagnetic when it has zero unpaired electrons. For each one we find the metal oxidation state, count its d electrons, note whether the ligand is strong or weak, and the geometry, then see if all electrons pair up.
Step 2: Test $[\text{Ni(CN)}_4]^{2-}$.
Here nickel is $+2$, which is $d^8$. Cyanide is a strong field ligand, so this four coordinate complex is square planar. In square planar $d^8$, all eight electrons pair into the four lower orbitals and the highest orbital stays empty. No unpaired electrons, so it is diamagnetic.
Step 3: Test $[\text{Co(CN)}_6]^{3-}$.
Cobalt is $+3$, which is $d^6$. Cyanide is a strong field ligand, so this is a low spin octahedral complex. In low spin $d^6$, all six electrons fill the lower three orbitals in pairs, leaving none unpaired. So this is diamagnetic too.
Step 4: Test $[\text{Fe(CN)}_6]^{3-}$.
Iron is $+3$, which is $d^5$. Even with strong field cyanide giving low spin, five electrons in three lower orbitals must leave one electron unpaired. So this is paramagnetic, not diamagnetic. Rejected.
Step 5: Test $[\text{Cu(H}_2\text{O)}_6]^{2+}$.
Copper is $+2$, which is $d^9$. Nine electrons can never fully pair, so there is always one unpaired electron. This is paramagnetic. Rejected.
Step 6: Collect the diamagnetic pair.
Only the nickel cyanide and the cobalt cyanide complexes have zero unpaired electrons.
\[ \boxed{[\text{Ni(CN)}_4]^{2-} \text{ and } [\text{Co(CN)}_6]^{3-} \text{ are diamagnetic}} \]