1. Output voltage. Both diodes are forward-biased by the 12 V source and V₀ is taken directly across them, so V₀ is one diode drop:
\(V_0 = V_D = 0.7\ \text{V}\) (silicon; 0.3 V for germanium).
It stays at 0.7 V whatever current is drawn, as long as the diodes keep conducting. The resistor current is
\(I_R = \dfrac{E - V_D}{R} = \dfrac{12 - 0.7}{330} \approx 34.2\ \text{mA}\).
With 20 mA drawn by the branch, the diodes still carry about 14.2 mA together, so they remain on and V₀ = 0.7 V.
2. Why two diodes in parallel. To share the current. A single diode would have to carry the whole ≈ 34 mA; if its rating is lower (e.g. 20 mA) it would overheat. Two identical diodes split it, roughly 17 mA each, keeping each within its rating. It also gives redundancy: if one opens, the other keeps the output clamped.
— rhs