Let x(t) = u(t-3) - u(t-5) and h(t) = e^-3tu(t)
A) y(t) = x(t)*h(t) I used the integral y(t) = $ \int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau $ for simplicity. Then, for t<3 y(t)=0 For 3 < t < 5 y(t) = (e^-9 - e^-3(t-3)) / 3 For t>5 y(t) = (e^-3(t-5) - e^-3(t-3)) / 3
