Step 1: (i)(I) Plus four:
Lanthanoids mostly give $Ln^{3+}$. Exceptions appear when the ion reaches an empty, half-filled or full 4f subshell.
Cerium is $[Xe]4f^1 5d^1 6s^2$. Remove all four outer electrons and the ion $Ce^{4+}$ has the noble gas core $[Xe]$.
Answer: Ce (Pr and Tb are other examples).
Step 2: (i)(II) Plus two:
Europium is $[Xe]4f^7 6s^2$. Only the two 6s electrons leave, so $Eu^{2+}$ keeps seven electrons in the 4f subshell.
A half-filled subshell is stable, so +2 is easy for europium.
Answer: Eu (Yb and Sm are other examples).
Step 3: (ii) Catalytic action:
A good catalyst must give the reaction an easier path. Transition metals do this in two ways.
Way 1: they have several oxidation states close in energy, so they can form an intermediate with a reactant and then return to the original state.
Way 2: their surface can hold reactant molecules by adsorption, and this brings the molecules together in the right position.
Both ways lower the activation energy. A familiar case is Ni in the hydrogenation of oils.
Step 4: (iii) Melting point of Cr and Mn:
Melting point depends on the strength of the metallic bond, which depends on the number of unpaired electrons that can bond.
Cr, $3d^5 4s^1$, has 6 unpaired electrons to share.
Mn, $3d^5 4s^2$, has a stable half-filled d subshell, so its d electrons take little part in bonding.
Cr therefore has stronger bonding and a higher melting point than Mn.
Final Answer:
Ce shows +4 and Eu shows +2. Transition metals are good catalysts because of variable oxidation states and surface adsorption. Cr melts higher because it has more unpaired electrons for metallic bonding.
\[ \boxed{\text{Ce (+4), Eu (+2)}} \]