Find increase in OH- ion concentration after base is added
The problem is that you are trying to subtract the values, whereas all you need to do is find an increase in ratio $\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1}$ knowing that $\mathrm{pOH} = - \log{[\ce{OH-}]}$ and $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$:

$$\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1} = \frac{10^{-\mathrm{pOH_2}}}{10^{-\mathrm{pOH_1}}} = 10^{\mathrm{pOH_1} - \mathrm{pOH_2}} = 10^{\mathrm{p}K_\mathrm{w} - \mathrm{pH_1} - \mathrm{p}K_\mathrm{w} + \mathrm{pH_2}} = 10^{\mathrm{pH_2} - \mathrm{pH_1}} = 10^{12 - 7} = 10^5$$