Finding new pH after NaOH added to buffer solution
Start by calculating $\mathrm{p}K_\mathrm{a}$ using the Henderson–Hasselbalch equation and the initial data given. $$\mathrm{p}K_\mathrm{a} = 3.35 - \log\frac{0.15}{0.2} = 3.47$$

Calculate the amount of substance of $\ce{HA}$ and $\ce{A- }$ after addition of $\ce{NaOH}$. $$n_{\ce{HA}} = 0.15\times0.5 - 0.0015 = 0.0765$$ $$n_{\ce{A- }} = 0.2\times0.5 + 0.0015 = 0.0985$$

Calculate the final $\mathrm{pH}$. The final volume is not needed because it is the same for both concentrations and so it cancels in the equation. $$\mathrm{pH} = 3.47 + \log\frac{0.0985}{0.0765} = 3.58$$