How to calculate the pH of a sodium bicarbonate-sodium hydroxide buffer solution?
Your idea is good, but you have thought of the wrong reaction.

You have written that $\ce{HCO_3^-}$ hydrolyses to give $\ce{H_2CO_3}$ and $\ce{ OH^-}$, but think carefully when you add base, $\ce{OH^-}$ is in excess in this medium. So, shouldn't your proposed reaction proceed in backward direction instead? Moreover, your reaction is actually not going to occur.

When you are adding strong base, $\ce{HCO_3^-}$ acts an acid to give $\ce{CO_3^2-}$ like:

$$\ce{NaHCO_3 + NaOH -&gt; Na_2CO_3 +H_2O}$$
 or, more simply
$$\ce{HCO_3^- + OH^- -&gt; CO_3^2- + H_2O}$$ 

Initially, number of millimoles of $\ce{HCO_3^-}$ were $=50\times0.05 = 2.5$ and number of millimoles of base added $=5\times0.1 = 0.5$

So, $\ce{CO_3^2-}$ produced will be also 0.5 millimoles as the base added was a limiting reagent, and $\ce{HCO_3^-}$ left $= 2.5-0.5 = 2$ millimoles.

As, you can see now, the solution acts an acid buffer as, $\ce{HCO_3^-}$ is a weak acid and $\ce{CO_3^2-}$ is the salt after reacting with a strong base. According to Henderson-Haselbach equation,

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{[\text{salt}]}/[\text{acid}]$$
$\mathrm{p}K_\mathrm{a}$ of $\ce{HCO_3^-}=\mathrm{p}K_{\mathrm{a2}}$ of $\ce{H_2CO_3^}=10.3$ . Now [$\ce{HCO_3^-}$] = $\pu{0.02 M}$, and [$\ce{CO_3^2-}$] = $\pu{0.005 M}$. Now if you calculate $$\mathrm{pH}= 10.3 +\log\frac{0.5}2 = 10.3- \log(4) = 9.69$$ So, the final $\ce{pH}$ of the medium has come out to be $= 9.69$.

The difference in digits after decimal may be due to some given $\mathrm{p}K_\mathrm{a2}$ which is experimental and you should do calculation according to that.