Part B: The basics of linearity
$ x_1(t) = e^{2jt} \rightarrow linear-system \rightarrow y_1(t) = te^{-2jt} $
$ x_2(t) = e^{-2jt} \rightarrow linear-system \rightarrow y_2(t) = te^{2jt} $
The input, cos(2t) is equal to $ \frac{1}{2}(e^{j2t} + e^{-j2t}) $
From the properties of a linear system $ ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t) $,
The response to cos(2t) is $ \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt} $
