Answer
a) $v_{max} = 13562.9 \mathrm{m/s}$. This occurs when the protons are infinitely far apart.
b) $a_{max} = 2.45\times10^{17} \mathrm{m/s^2}$. This occurs when the protons are closest together.
Work Step by Step
The maximum velocity they will obtain is when they are infinitely far apart and all the electrostatic potential energy is converted into kinetic energy. From energy conservation,
$\frac{e^2}{4\pi\epsilon_0 r} = 2\times \frac{1}{2}m_Pv_{max}^2$
where $e = 1.6\times10^{-19}\mathrm{C}$ is the charge of the proton, $r = 0.75 \mathrm{nm} = 7.5\times 10^{-10}\mathrm{m}$ is the initial distance and $m_P = 1.67\times 10^{-27} \mathrm{kg}$ is the mass of a proton.
Thus by solving for velocity, we find that the maximum velocity attained by each proton is
$v_{max} = \sqrt{\frac{e^2}{4\pi\epsilon_0 rm_P}} = \sqrt{\frac{9\times10^9\times( 1.6\times10^{-19})^2}{7.5\times 10^{-10}\times1.67\times 10^{-27}}}\mathrm{m/s} = 13562.9 \mathrm{m/s}$.
The maximum acceleration is when the particles are closest together and has the value
$a_{max} = F_{max}/m_p =\frac{e^2}{4\pi\epsilon_0 r^2m_P} = \frac{9\times10^9\times( 1.6\times10^{-19})^2}{(7.5\times 10^{-10})^2\times1.67\times 10^{-27}}\mathrm{m/s^2} = 2.45\times10^{17} \mathrm{m/s^2}$.
Even though this value is quite high, the acceleration drops drastically as the distance between the protons increases. The speed of the protons thus never exceeds the speed of light, which is not allowed.
