We wish to compute
$$\int_{-\infty}^{\infty} dx \frac{x}{(1+x^2)^2} e^{i k x} = -i \frac{d}{dk}\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^2} e^{i k x}$$
We know from the linked-to result that
$$\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^2} e^{i k x} = \frac{\pi}{2}(1+|k|) e^{-|k|}$$
Thus for $k \gt 0$ we have
$$-i \frac{d}{dk}\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^2} e^{i k x} = i \frac{\pi}{2} k \, e^{-k}$$
and for $k \lt 0$ we have
$$-i \frac{d}{dk}\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^2} e^{i k x} = i \frac{\pi}{2} k \, e^{k}$$
Therefore, the FT of $x/(1+x^2)^2$ is
$$\int_{-\infty}^{\infty} dx \frac{x}{(1+x^2)^2} e^{i k x} = i \frac{\pi}{2} k \, e^{-|k|}$$
To compute the FT of $1/(1+i x)^2$, consider the contour integral
$$\oint_C \frac{dz}{(1+i z)^2} e^{i k z}$$
where $C$ is a semicircle of radius $R$ in the upper half plane for $k \gt 0$ and in the lower half plane for $k \lt 0$. Note that there is a double pole at $z=i$ and no other poles elsewhere; thus, the FT is zero when $k \lt 0$. When $k \gt 0$, however, we evaluate the contour integral via the residue theorem. Taking the limit as $R \to\infty$, we have
$$\begin{align}\int_{-\infty}^{\infty} \frac{dx}{(1+i x)^2} e^{i k x} &= i 2 \pi \operatorname*{Res}_{z=i} \frac{e^{i k z}}{(1+i z)^2}\\ &= -i 2 \pi \left [\frac{d}{dz} e^{i k z} \right ]_{z=i}\\ &= 2 \pi k \, e^{-k} \end{align}$$
Thus
$$\int_{-\infty}^{\infty} \frac{dx}{(1+i x)^2} e^{i k x} = 2 \pi k\, e^{-k} \, \theta(k)$$
Computing the FT of $\cos{(\pi x/2)}$ involves deforming the above contours about the "poles" at $z=\pm 1$, even though the singularities are really removable here. Thus we will be dealing with Cauchy principal values, but only temporarily, as you will see below.
Consider the following contour integral
$$\oint_C dz \frac{e^{i (k+\pi/2) z}}{1-z^2}$$
where $C$ is, for $k+\pi/2 \gt 0$, a semicircle in the upper half plane of radius $R$, but with semicircular deformities of radius $\epsilon$ above the real line above the poles at $z=\pm 1$. For $k+\pi/2 \lt 0$, we use the lower half plane with analogous deformities at the poles.
Thus for $k+\pi/2 \gt 0$, the contour integral is
$$\int_{-R}^{-1-\epsilon} dx \frac{e^{i (k+\pi/2) x}}{1-x^2} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{ e^{i (k+\pi/2) (-1+\epsilon e^{i \phi})}}{1-(-1+\epsilon e^{i \phi})^2} \\ + \int_{-1+\epsilon}^{1-\epsilon} dx \frac{e^{i (k+\pi/2) x}}{1-x^2} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{ e^{i (k+\pi/2) (1+\epsilon e^{i \phi})}}{1-(1+\epsilon e^{i \phi})^2} \\+ \int_{1+\epsilon}^R dx \frac{e^{i (k+\pi/2) x}}{1-x^2}+ i R \int_0^{\pi} d\theta\, e^{i \theta} \frac{ e^{i (k+\pi/2) R e^{i \theta}}}{1-R^2 e^{i 2 \theta}} $$
We can show that the sixth integral vanishes as $R\to\infty$, as the magnitude of that integral is bounded by
$$\frac{2}{R} \int_0^{\pi/2} d\theta \, e^{-(k+\pi/2) R \sin{\theta}} \le \frac{2}{R} \int_0^{\pi/2} d\theta \, e^{-2(k+\pi/2) R \theta/\pi} \le \frac{\pi}{R^2 (k+\pi/2)}$$
The 1st, 3rd, and 5th integrals combine to form the Cauchy PV of the original FT, and we consider the 2nd and 4th integrals in the limit as $\epsilon \to 0$.
By Cauchy's Integral Theorem, because there are no poles in the interior of $C$, the contour integral is zero. Thus, we have in the limits as $R\to\infty$ and $\epsilon \to 0$:
$$PV \int_{-\infty}^{\infty} dx \frac{e^{i (k+\pi/2) x}}{1-x^2} = i \frac{\pi}{2} (e^{-i (k+\pi/2)} - e^{i (k+\pi/2)}) = \pi \cos{k}$$
Note that this is valid only for $k+\pi/2 \gt 0$. For $k+\pi/2 \lt 0$, we use the contour in the lower half plane and make similar calculations. The result is that
$$PV \int_{-\infty}^{\infty} dx \frac{e^{i (k+\pi/2) x}}{1-x^2} = \pi \cos{k} \,\operatorname*{sgn}{\left ( k+\frac{\pi}{2}\right)}$$
We find, similarly, that
$$PV \int_{-\infty}^{\infty} dx \frac{e^{i (k-\pi/2) x}}{1-x^2} = -\pi \cos{k} \,\operatorname*{sgn}{\left ( k-\frac{\pi}{2}\right)}$$
Combining these two expressions, we finally get that
$$ \int_{-\infty}^{\infty} dx \frac{\cos{(\pi x/2)}}{1-x^2} e^{i k x} = \frac{\pi}{2} \cos{k} \left [ \operatorname*{sgn}{\left ( k+\frac{\pi}{2}\right)} - \operatorname*{sgn}{\left ( k-\frac{\pi}{2}\right)} \right ]$$
or,
$$ \int_{-\infty}^{\infty} dx \frac{\cos{(\pi x/2)}}{1-x^2} e^{i k x} = \begin{cases} \pi \cos{k} & |k| \lt \frac{\pi}{2} \\ 0 & |k| \gt \frac{\pi}{2} \end{cases}$$
