Answer
(a) No
(b) $0.98c$
Work Step by Step
(a) The inventor is using simple velocity addition which is valid only for $v\lt \lt c$. Thus, it is not advisable to invest in this scheme.
(b) We know that
$v=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
We plug in the known values to obtain:
$v=\frac{0.80c+0.80c}{1+\frac{(0.80c)(0.80c)}{c^2}}$
$v=0.98c$
