Adjusting the ph of salicylic acid using sodium hydroxide
The Henderson-Hasselbalch equation is the correct equation, but ask yourself what you are putting in there:
$$pH=pK_a + \log\left(\frac{c(A^-)}{c(AH)}\right)$$

Since $A^-$ and $HA$ occupy the same volume, you can rewrite:
$$pH=pK_a + \log\left(\frac{n(A^-)}{n(AH)}\right)$$

When adding Sodium Hydroxide consider the following equation
$$\ce{AH + NaOH &lt;=&gt; A- + H2O + Na+}$$

Given that you will find 
$$ n(A^-) = n(\ce{NaOH})\\
   n(HA) = n_0(HA)-n(\ce{NaOH})$$

Then you can substitute and solve for $n(\ce{NaOH})$:
$$pH=pK_a + \log\left(\frac{n(\ce{NaOH})}{n_0(HA)-n(\ce{NaOH})}\right)$$

You then divide by the concentration, yielding your volume.



Intermediate results: 

Initial amount of $HA$:


       $n_0(HA) = 0.5825\ \text{mmol}$


Amount of $\ce{NaOH}$:


   $n(\ce{NaOH}) \approx 0.4993\ \text{mmol}$


Final Volume:


   $V(\ce{NaOH}) = 3.22\ \text{mL}\ \text{(3sf)}$