Part A
Define $ y(t)=\int_{-\infty}^{\infty}\frac{1}{2}x(t)dt $.
h(t) is the system response of $ x(t)=\delta (t) $.
$ h(t)=\int_{-\infty}^{\infty}\frac{1}{2}\delta (t)dt=\frac{1}{2}u(t) $.
$ H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}=\int_{-\infty}^{\infty}\frac{1}{2}u(\tau)e^{-s\tau} $
$ H(s)=\int_{0}^{\infty}\frac{1}{2}e^{-s\tau}=\frac{-1}{2s}e^{-s\tau}|_0^\infty=\frac{-1}{2s}(e^{-\infty}-e^{0}) $
$ H(s)=\frac{1}{2s} $
Part B
When $ x(t) = \sum^{\infty}_{k = -\infty} a_ke^{jK{w_0}t}\!= $,
the response y(t) is equal to$ \sum^{\infty}_{k = -\infty} a_kH(jw)e^{jK{w_0}t}\! $
From Question 1:
$ x(t) = 4cos(4t)+(2+12j)sin(12t)=2e^{1(j4t)}+2e^{-1(j4t)}+(6-j)e^{3(j4t)}+(j-6)e^{-3(j4t)} $
$ y(t) = \sum^{\infty}_{k = -\infty} a_kH(jw)e^{jK{w_0}t}\!=2H(j4)e^{1(j4t)}+2H(j4)e^{-1(j4t)}+(6-j)H(j4)e^{3(j4t)}+(j-6)H(j4)e^{-3(j4t)} $
$ y(t)=\frac{-1}{4}je^{1(j4t)}+\frac{-1}{4}je^{-1(j4t)}+(\frac{-1}{8}-\frac{3}{4}j)e^{3(j4t)}+(\frac{1}{8}+\frac{3}{4}j)e^{-3(j4t)} $
