The shift is defined by $g_a(x) = f(x-a)$. Then you write
$$F[g_a](\xi) = \int_{\Bbb R}g_a(x)\exp(-ix\xi)dx = \int_{\Bbb R}f(x-a)\exp(-ix\xi)dx.$$
Conceptually, you first apply the shift and then apply the Fourier transform, but you can apply the shift only to the function, there is no sense in applying it to the exponent.
NOTE: I would advise you not to you write $\mathcal F[h(x)]$ for the Fourier transform of a function $h(x)$, as when you calculate the Fourier transform you integrate over $x$. It's better to write, for example $\mathcal F[h](k)$, instead.
I think there is a sign missing when calculating the Fourier transform of $\sin \alpha x$, and the Fourier transform of $h$ should be
$$\mathcal F[h](k) = -\frac{i}{1+k^2}(\delta(a-k)-\delta(a+k))$$
the Fourier transform of $h$. The inverse Fourier transform (I might have put the wrong constant in front as I am not sure what normalisation you are using: you can easily fix that) gives you back $h$ as
$$ h(x) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \hat h(k) e^{+ i kx}dk =-\frac{i}{1+a^2}\left( e^{i\alpha x}-e^{-i\alpha x}\right) = \frac{2}{1+\alpha^2} \sin (\alpha x). $$
So your constant $A$ would be equal to $2/(1+\alpha^2)$.
For the last part, call $f(x) = \sin \alpha x$ and $g(x) = e^{-|x|}$. Then, Since
$$ \mathcal F[h] = \mathcal F[f] \mathcal F[g], $$
you have
$$ h = \mathcal F^{-1} \left( \mathcal F[f] \mathcal F[g]\right) = \left(\mathcal F^{-1}\mathcal F\right)[f] * \left(\mathcal F^{-1}\mathcal F\right)[g]= f * g, $$
where $*$ denote the convolution product. In other words, we get
$$\frac{2 \sin (\alpha x)}{1+\alpha^2} = h(x)= (f*g)(x) =\int_{-\infty}^{+\infty} \sin(\alpha p) e^{-|x-p|} dp,$$
which holds for all $\alpha$, which you may as well call $\beta$.
Best Answer
It is easy to show that the following Fourier integral
$$F(\omega)=\int_{-\infty}^{\infty}dx e^{-zx^2-i\omega x}$$
converges iff $\Re(z)>0.$ We can perform this integral for any $z$ satisfying this condition by completing the square and performing an appropriate change of contour the in the complex $x$ plane and we get the result
$$F(\omega)=\sqrt{\frac{\pi}{z}}e^{-\omega^2/4z}$$
Now, it is clear that $z=-i\alpha$ does not satisfy $\Re(z)>0$- however the integral converges for arbitrarily small real parts $\Re(z)=\epsilon>0$ and can be shown to approach a limiting value as $\epsilon\to 0, \Im(z)\neq 0$. Since the integral can be shown to converge for $\Re(z)=0, \Im(z)\neq 0$ and is continuously connected to the sector $\Re(z)>0, \Im(z)\neq 0$ we can conclude by taking the appropriate limit that
$$\int_{-\infty}^\infty e^{i\alpha x^2-i\omega x}dx=\begin{Bmatrix}\sqrt{\frac{\pi}{\alpha}}e^{i\omega^2/4\alpha+i\pi/4}&, ~~\alpha\neq 0\\2\pi\delta(\omega)&, ~~\alpha=0\end{Bmatrix}$$