How do I calculate how much sodium hydroxide I need to add to get a new pH?
At the beginning, the amount of substance of hydronium ions is: 
$$n(\ce{H3O+})=10^{-3.7}  \times 100 \times10^{-3}=10^{-4.7} \mathrm {mol}$$
After adding sodium hydroxide, the amount of substance of hydronium ions is: 
$$n(\ce{H3O+})=10^{-4.5}  \times 100 \times10^{-3}=10^{-5.5} \mathrm {mol}$$ 

The difference represents the amount of substance of hydronium ions that have reacted with hydroxide ions.
$$n(\ce{H3O+})=10^{-4.7}-10^{-5.5}= n(\ce{OH-})$$
The volume of sodium hydroxide to add is:
$$v(\ce{OH-})=\frac {10^{-4.7}-10^{-5.5}}{10}= 1.68\times10^{-6}\mathrm{L}=1.68\,\mu \mathrm{L} $$

Assuming the volume of a droplet to be about 60 micro liter, it is much less than a droplet!