Re: convolution: x(t) = u(t)-2u(t-2)+u(t-5), h(t) = e^(2t) u
\(\displaystyle \L (x*h) (t) = \int_0^t x(s) h(t-s) ds\)
\(\displaystyle \L = \int_0^t \left( u(s)-2u(s-2)+u(s-5) \right) \left( e^{2s} u(1-s) \right) ds\)
It looks complicated if you attempt to evaluate this integral. Since I see the step function u there, I bet you are expected to use Laplace transforms. Remember,
\(\displaystyle \L \mathcal{L} ( x * h) (s) = X(s) H(s)\)
where X and H are the transforms of x and h. You can find those easily, multiply, and use inverse Laplace transform to get x*h.