Step 1: Anchor on the default oxidation state.
Every lanthanide's most stable, characteristic state is +3.
Step 2: Think about what a reducing agent needs to do.
A species that sits below +3 wants to lose an electron and climb up to that stable +3 state, and losing an electron is exactly what makes it a reducing agent.
Step 3: Scan the options with that lens.
$\text{Eu}^{2+}$ and $\text{Yb}^{2+}$ both sit below +3, so they readily get oxidised to $\text{Eu}^{3+}$/$\text{Yb}^{3+}$, acting as reducing agents. $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$ sit above +3 and instead want to gain an electron (oxidising agents), while $\text{Gd}^{3+}$, $\text{Lu}^{3+}$, $\text{La}^{3+}$, $\text{Pm}^{3+}$ are already at the comfortable +3 state and show no such urge.
Final answer: Option 2, $\text{Eu}^{2+}, \text{Yb}^{2+}$.