The Basics of Linearity
The problem tells us that:
$ \ e^{2jt} \longrightarrow sys \longrightarrow t e^{-2jt} $ and that $ \ e^{-2jt} \longrightarrow sys \longrightarrow t e^{2jt} $
and asks us to find the output for an input of $ \ \cos{2t} $. However due to Euler Identity, we can write $ \ \cos{2t} $ as a linear combination of the two exponential inputs above. Then as a result of the systems linearity, the output will be the just be the linear combination of the outputs of the two exponential above.
$ \ \cos{2t}= 0.5 e^{2jt} +(- 0.5 e^{-2jt}) $
Therefore putting this into the system will yield:
$ \ \cos{2t}= 0.5 e^{2jt} +(- 0.5 e^{-2jt}) \longrightarrow sys \longrightarrow t(- 0.5 e^{-2jt})+ t 0.5 e^{2jt} = t(0.5 e^{2jt} +(- 0.5 e^{-2jt})) $
$ \ = t \cos(2t) $
