Step 1: Recall IUPAC rules for naming coordination compounds.
(i) Name cation before anion. (ii) Within complex ion, name ligands alphabetically before the metal. (iii) Anionic ligands end in "-o" (chlorido, fluorido). (iv) Neutral ligands use special names (en = ethane-1,2-diamine, water = aqua, $NH_3$ = ammine). (v) Metal oxidation state in Roman numerals. (vi) Anionic complex ions: metal name + "-ate".
Step 2: Name compound (i): $[PtCl_2(en)_2]SO_4$.
Cation: $[PtCl_2(en)_2]^{2+}$. Anion: $SO_4^{2-}$ (sulphate). Ligands: $Cl^-$ = chlorido (x2, prefix: di-); $en$ = ethane-1,2-diamine (x2, prefix for polydentate: bis-). Alphabetical: c before e, so chlorido first. Oxidation state of Pt: $x + 2(-1) + 2(0) = +2 \Rightarrow x = +4$. IUPAC name: Dichloridо bis(ethane-1,2-diamine)platinum(IV) sulphate.
Step 3: Write the name of (i) clearly.
Dichloridо bis(ethane-1,2-diamine)platinum(IV) sulphate.
Step 4: Name compound (ii): $(NH_4)_2[CoF_4]$.
Cation: $NH_4^+$ (ammonium, x2). Complex anion: $[CoF_4]^{2-}$. Ligand: $F^-$ = fluorido (x4 = tetrafluorido). Oxidation state of Co: $x + 4(-1) = -2 \Rightarrow x = +2$. Complex is anionic, so metal gets "-ate": cobalt becomes cobaltate. IUPAC name: Ammonium tetrafluoridocobaltate(II).
Step 5: Verify both oxidation states.
(i) $Pt^{4+}$ + $2Cl^-$ + $2(en)^0$ gives cation charge $+4-2 = +2$. One $SO_4^{2-}$ gives $-2$. Neutral overall. (ii) $Co^{2+}$ + $4F^-$ gives anion charge $+2-4 = -2$. Two $NH_4^+$ give $+2$. Neutral overall. Both verified.
Step 6: State the final IUPAC names.
(i) Dichlorido bis(ethane-1,2-diamine)platinum(IV) sulphate. (ii) Ammonium tetrafluoridocobaltate(II). \[ \boxed{(i)\;\text{Dichlorido bis(ethane-1,2-diamine)platinum(IV) sulphate};\;(ii)\;\text{Ammonium tetrafluoridocobaltate(II)}} \]