Part A
$ y(t) = K x(t-a) $
if $ x(t)=e^{jwt} $ was inputed to the system
$ y(t) = K e^{jw(t-a)} $
$ = K e^{-jwa}e^{jwt} $
eigen function is $ e^{-jwa} $
$ H(jw)=Ke^{-jwa} $
$ h(t)=K\delta (t-a) $
$ H(s)=\int_{-\infty}^{\infty}K\delta (\tau -a)e^{-s\tau}d\tau=Ke^{-as} $
