What we know
The two things that are important when solving this problem is that we know what the output is for $ e^{j2t} $ and $ e^{-j2t} $ and that the system is linear.
$ e^{j2t} $ ---> system ---> $ te^{-j2t} $
$ e^{-j2t} $ ---> system ---> $ te^{j2t} $
What we can do with this
First step is to take the input $ cos(2t) $ and relate it to the two known inputs. It is easy to do this using Euler's Identity $ e^{jat} = \frac{cos(t) + jsin(t)}{a} $
$ cos(2t) = \frac{cos(2t) + jsin(2t)}{2} + \frac{cos(2t) - jsin(2t)}{2} $
