How to apply the Fourier Transform to this?
As you mentioned:
$$\operatorname{FT}(\cos(t)) = \pi[\delta (\omega-1) + \delta (\omega+1)]$$
$$\operatorname{FT}(e^{-3t}u(t))=\frac{1}{3+j\omega}$$

The multiplication in the time domain is the convolution in the frequency domain with factor \$\frac{1}{2\pi}\$:
$$\frac{1}{2\pi}\pi[\delta (\omega-1) + \delta (\omega+1)] * \left(\frac{1}{3+j\omega}\right)=$$
$$=\frac{1}{2}[\delta (\omega-1) + \delta (\omega+1)] * \left(\frac{1}{3+j\omega}\right)=$$
As convolution of a function \$f(\omega)\$ with \$\delta (\omega-a)\$ is \$f(\omega-a)\$:

$$=\frac{1}{2} \left(\frac{1}{3+j(\omega-1)} + \frac{1}{3+j(\omega+1)}\right)=$$

$$=\frac{1}{2}\left(  \frac{3+j(\omega+1)+3+j(\omega-1)}{(3+j(\omega+1))(3+j(\omega-1))}  \right)=$$
$$=\frac{1}{2}\left( \frac{6+ 2j\omega}{ 9+3j(\omega+1+\omega-1)-(\omega+1)(\omega-1) }\right)=$$
$$=\frac{3+ j\omega}{ 9+6j\omega-\omega^2+1 }=$$
$$=\frac{3+ j\omega}{ (3+j\omega)^2+1 }$$
Add the factor \$5\$ we omitted in the beginning, and you will get your result.