Given the 3 following: $$\mathfrak{F}(e^{-|t|})=\sqrt\frac{2}{\pi}\frac{1}{1+\omega^2}$$ $$\mathfrak{F}(r(t))=\sqrt\frac{2}{\pi}\frac{\sin \omega}{\omega}$$ where $$r(t)=\left\{\begin{matrix} 1, &\left | t \right |<1\\ 0, &\left | t \right |>1 \end{matrix}\right.$$ I can say that if $f(t)={\mathfrak{F}}^{-1}\left ( \frac{\sin \omega}{\omega(1+\omega^2)} \right )$ then $$f(t)=\sqrt\frac{\pi}{2}r(t)*\sqrt\frac{\pi}{2}e^{-|t|}$$ or $$f(t)=\frac{\pi}{2}r(t)*e^{-|t|}$$ Now I want to find $f(0)$ and $f(1)$. How could this be done?
And, what is the relation between $f(\omega)$ and $\mathfrak{F}\left(\frac{\sin t}{t(1+t^2)} \right)$ [if there is any].
You should write it as $f(t) = \frac\pi2 (r \ast e^{-|\cdot|})(t)$ to not reuse $t$. These two things are just two definite convolution integrals:
$$f(0) = \int_{-1}^1 e^{-|t|} \mathrm dt\\ f(1) = \int_{-1}^1 e^{-|t-1|} \mathrm dt$$
Regarding the last question, know that $\mathfrak F(\mathfrak F(f))(t) = f(-t)$ for any $f\in L^1(\mathbb R) \cap L^2(\mathbb R)$