How do I find the final concentration after an additional dilution?
Check your math: It looks like you forgot to divide by $ 10\ \mathrm{ml} $ in finding $ c_2 $. 

Also: the process you are describing in the first paragraph will not produced the same concentration that you started with. If you want $ 35\ \mathrm{ml} $ at the same concentration ($ 0.02\ \mathrm{g/ml} $) by taking some fraction of your original solution and diluting it, you will need to add more solute.

e.g.
$$ \rm(0.02\ g/ml) \cdot (35\ ml) = 0.07\ g $$
$$ \rm(0.07\ g + \mathit{m}) / 35\ ml = (0.02\ g/ml) $$
$$ m \rm = (0.02\ g/ml) \cdot (35\ ml) - 0.07\ g $$

The first expression is the concentration of your fraction (which doesn't change from the original solution, assuming a homogenous solution).

The second expression is equating the concentration that you want (conc. of the original solution, for example, with the concentration of your diluted fraction with $ m\ \mathrm{g} $ of solute added (using the mass we just found).

Then I rearranged this to express the mass of solute needed to obtain this concentration.