Calculating output of a system given impulse response and input
Eventhough you could solve this problem using other means, such as frequency domain methods, you could also follow a direct time domain path as the following.
Given the impulse response $h(t) = e^{-t} u(t)$ of an LTI system and the applied excitation input $x(t) = \cos(2\pi t)$ (which is of infinite extend from $t=-\infty$ to $t=\infty$), the output can be written as the convolution integral:
$$y(t) = x(t)\star h(t) = \int_{-\infty}^{\infty} h(t-\tau) x(\tau)d\tau = \int_{-\infty}^{\infty} e^{-(t-\tau)}u(t-\tau) \cos(2\pi \tau)d\tau$$
$$y(t)  = e^{-t} \int_{-\infty}^{t} e^{\tau} \cos(2\pi \tau)d\tau$$
At this point we have to evaluate the integral which can be found on most calculus texts or in an integral table. I will use MATLAB to evaluate it which results in
$$\boxed{ y(t) =  \frac{\cos(2\pi t) }{1 + 4\pi^2} + \frac{2\pi \sin(2\pi t) }{1 + 4\pi^2} }$$
You can simplify the result using $a\cos(wt) + b\sin(wt) = \sqrt{a^2+b^2} \cos(wt - \tan^{-1}(b/a) )$ into
$$\boxed{ y(t) =  \frac{1 }{\sqrt{1 + 4\pi^2}} \cos(2\pi t)}$$
as the solution. The solution has no transient part as the input was applied beginning from $t=-\infty$.