How to calculate the pH of the neutralisation of HCN with excess of KOH?
While you've generally done a good job, you've made an important error. The hydrolysis of the cyanide ion should be written as an equilibrium reaction:

$$\ce{CN-(aq) + H2O(l) &lt;=&gt;  HCN(aq) + OH-(aq)}$$

That's because $\ce{HCN}$ is a very weak acid and its conjugated base a weak base (homework: look up the $K_b$ of $\ce{CN-}$).

Now, as you duly noted, there's excess $\ce{KOH}$ in the solution, which dissociates completely. The $\ce{OH-}$ resulting from that dissociation now push the equilibrium:

$$K_b=\frac{[\ce{HCN}]\times[\ce{OH-}]}{[\ce{CN-}]}$$

to the left. As a result, the contribution of the $[\ce{OH-}]$ coming from the hydrolysis of the cyanide is negligible and can be ignored. Only the $[\ce{OH-}]$ from the excess $\ce{KOH}$ is to be counted. So $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$.

We can confirm that numerically by assuming $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$ and with:

$$\frac{[\ce{HCN}]}{[\ce{CN-}]}=\frac{K_b}{[\ce{OH-}]}\approx 0.0027$$

So neglibibly little cyanide is present as the acid $\ce{HCN}$.



As regards the matter of adding molarities, in some instances that is allowed.

Say we add $n_1$ moles of $X$ to $V\ \mathrm{L}$, that would give a molarity $M_1=\frac{n_1}{V}$. At a later stage we add $n_2$ moles of $X$, that molarity would be $M_2=\frac{n_2}{V}$.

The total molarity would be:

$$M=M_1+M_2=\frac{n_1}{V}+\frac{n_2}{V}=\frac{n_1+n_2}{V}$$

But if we were to mix volumes of solutions it would be:

$$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$$

Now  they are no longer simply additive.