Step 1: Recall the naming rules.
Name the cation first and then the anion. Inside a complex ion, name ligands in alphabetical order with number prefixes, then the metal, and show the metal's oxidation state in Roman numerals. The metal in an anionic complex ends in -ate.
Step 2: Work out part (i) cation \([Ag(NH_3)_2]^+\).
Two \(NH_3\) (ammine) ligands are neutral, so silver is in the +1 state. This cation is diamminesilver(I).
Step 3: Work out part (i) anion \([Ag(CN)_2]^-\).
Two \(CN^-\) (cyano) ligands carry -2 charge total and the ion charge is -1, so silver is +1. Because this is an anionic complex, silver becomes argentate, giving dicyanidoargentate(I).
Step 4: Combine part (i).
The full name is diamminesilver(I) dicyanidoargentate(I).
Step 5: Work out part (ii) \(K_3[Fe(C_2O_4)_3]\).
Three oxalate \((C_2O_4^{2-})\) ligands give -6, and three \(K^+\) give +3, so iron is +3. The oxalato ligands are named first, the complex is anionic so iron becomes ferrate, giving potassium trioxalatoferrate(III).
Answer: (i) diamminesilver(I) dicyanidoargentate(I); (ii) potassium trioxalatoferrate(III).