what happens when a small amount of NaOH(aq) is added to H2CO3(aq)+H2O -> H3O+(aq) + HCO3-(aq)?

Small amount of NaOH....

H2CO3(aq) + H2O(l) --> H3O+(aq) + HCO3^-(aq)

First problem. There is no molecular H2CO3 in aqueous solution. It doesn't exist in water. What is called "carbonic acid" is a solution of CO2 in equilibrium with small amounts of H+ and HCO3^-.
..... CO2(aq) + HOH(l) <==> H+ + HCO3^- ........... plus a very small amount of CO3^2-

Adding a small amount of NaOH(aq) will provide OH- to react with both H+ and HCO3- and shift the equilibrium to the right, decreasing the amount of dissolved CO2 and increasing the concentration of CO3^2-.

Generally, a small amount of NaOH(aq) is added to an aqueous solution, the pH would rise a lot. However, the given solution is a buffer solution, and thus the solution resists to change its pH when a small amount of NaOH(aq) is added.

H₂CO₃(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HCO₃⁻(aq) …… Kₐ₁

Before addition of NaOH, apply Henderson-Hasselbalch equation:
Original pH = pKₐ₁ + log{[HCO₃⁻]ₒ/[H₂CO₃]ₒ}

After addition of a small amount of H₂CO₃ is converted to of HCO₃⁻ ion. According to Le Chatelier principle, the equilibrium position would shift to the left and some HCO₃⁻ is converted back to H₂CO₃. As a result, only a very small amount of H₂CO₃ (say, Δx M) is converted to HCO₃⁻ ion.
Apply Henderson-Hasselbalch equation again:
New pH = pKₐ₁ + log{([HCO₃⁻]ₒ + Δx)/([H₂CO₃]ₒ - Δx)}

As [HCO₃⁻]ₒ ≫ Δx and [H₂CO₃]ₒ ≫ Δx, Δx is negligible.
Hence, log{([HCO₃⁻]ₒ + Δx)/([H₂CO₃]ₒ - Δx)} ≈ log{[HCO₃⁻]ₒ/[H₂CO₃]ₒ}
and thus (New pH) ≈ (Original pH) when a small amount of NaOH(aq) is added.