jmc

We can simplify the congruence as follows: $$\begin{array}{rl}64x& \equiv 2(\mathrm{mod}66)\\ 32x& \equiv 1(\mathrm{mod}33)\\ -x& \equiv 1(\mathrm{mod}33)\\ x& \equiv -1(\mathrm{mod}33)\\ x& \equiv 32(\mathrm{mod}33).\end{array}$$ The first few positive solutions to this are $32$, $32+33=65$, $32+2\cdot 33=98$, after which the solutions are clearly greater than $100$ and so are extraneous. Thus there are $\boxed{3}$ solutions in the given range.