Answer
a). 3200GeV
b). 38.7 GeV
Work Step by Step
a). since $E_{a}$ is much larger than proton rest energy,
we use $E_{a}^{2}=2mc^{2}\times E_{m}$
Hence, $E_{m}=3200GeV$
b). for colliding beams, total momentum=0.
Available energy $E_{a}$ is total energy of 2 colliding particles.
For proton-proton collision, each beam have same energy, so energy of each beam = $\frac{1}{2} E_{a}=38.7GeV.$
