The basics of linearity
$ cos(2t)=\frac{1}{2}(e^{2jt}+e^{-2jt}) \! $
We are given that the system's output for $ e^{2jt} \! $ is $ t*e^{-2jt} \! $, and the output for $ e^{-2jt} \! $ is $ t*e^{2jt} \! $. Since the system is linear, we can say that $ \frac{1}{2}(te^{-2jt}+te^{2jt})=tcos(2t) $.
It is rather straightforward.
