Write $\sin{x} = (e^{i x}-e^{-i x})/(2 i)$. Then consider the integral
$$PV \oint_{C_{\pm}} dx \frac{e^{\pm i z}}{z (z^2-2 z+2)} $$
where $C_{\pm}$ is a semicircular contour of radius $R$ in the upper/lower half plane with a semicircular detour into the upper/lower half plane of radius $\epsilon$. For $C_{+}$, we have
$$PV \oint_{C_{+}} dz \frac{e^{i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta}}}{R e^{i \theta} (R^2 e^{i 2 \theta} - 2 R e^{i \theta}+2)} $$
For $C_-$, we have
$$PV \oint_{C_{-}} dz \frac{e^{-i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{-i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{-\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{-i x}}{x (x^2-2 x+2)}- i R \int_0^{\pi} d\theta \, e^{-i \theta} \frac{e^{-i R e^{-i \theta}}}{R e^{-i \theta} (R^2 e^{-i 2 \theta} - 2 R e^{-i \theta}+2)} $$
In both cases, we take the limits as $R \to \infty$ and $\epsilon \to 0$. Note that, in both cases, the respective fourth integrals have a magnitude bounded by
$$\frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi}\le \frac{\pi}{R^3}$$
The respective second integrals of $C_{\pm}$, on the other hand, become equal to $\mp i \frac{\pi}{2} $. Thus,
$$PV \oint_{C_{\pm}} dz \frac{e^{\pm i z}}{z (z^2-2 z+2)} = PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2}$$
On the other hand, the respective contour integrals are each equal to $\pm i 2 \pi$ times the sum of the residues of the poles inside their contours. (For $C_-$, there is a negative sign because the contour was traversed in a clockwise direction.) The poles of the denominator are at $z_{\pm}=1 \pm i$. Thus,
$$PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2} = \pm i 2 \pi \frac{e^{\pm i (1 \pm i)}}{(1 \pm i) (2) (\pm i)} $$
Taking the difference between the two results and dividing by $2 i$, we get that
$$\int_{-\infty}^{\infty} dx \frac{\sin{x}}{x (x^2-2 x+2)} = \frac{\pi}{2} \left (1+\frac{\sin{1}-\cos{1}}{e} \right ) $$
Note that we may drop the $PV$ because the difference between the integrals removes the pole at the origin.
It's true that you can shift the poles instead of deforming the contour, but you only get the same result if the direction in which you shift the poles corresponds to the direction in which you deform the contour. See Wikipedia for different ways of interpreting this integral by choosing different semicircles around the poles, corresponding to different shifts in the denominator.
If you just write the integral like that and don't specify what you mean, it's not well-defined, and you can get any of the three results you quoted by interpreting it differently. Usually, without further specification, one would consider the principal value, which effectively takes the average of the upper and lower semicircles at each pole, thus gets half a contribution from each pole and thus yields the middle one of the three results you quoted.
Best Answer
Note that in order to apply the residue theorem, it is not enough that the function is analytic on the original contour, it has to be analytic on the full contour and its interior, except possibly at a finite number of points.
If we restrict to $x > 0$ we have an analytic function, which can be extended to a meromorphic function
$$f_+(z) = \frac{z^{1/2}}{(z - 4i)(z + 2i)}$$
on the whole plane, but it doesn't coincide with the integrand on the negative real axis. There the integrand can also be extended to a meromorphic function, namely
$$f_-(z) = \frac{(-z)^{1/2}}{(z - 4i)(z + 2i)}.$$
The integral we are interested in, is the sum of the integral of $f_+$ over the positive real axis, and $f_-$ over the negative real axis. We will compute these in two steps.
We turn the positive real axis and the negative real axis into two different closed contours as in these images:
The red line indicates a branch cut, and the branch of the functions has to be fixed in such a way that it coincides with the integrand on the upper edge of the branch cut.
It is not hard to see that in both cases, the integral over the lower edge is equal to that over the upper edge: the square root picks up a minus sign, but the integral is taken in the opposite direction. It is not hard to see either that the integrals over both the small and the big circle segments vanish in the limit, and it follows that the integral over each of the contours is exactly twice the value of the integral we're interested in.
We compute these integrals using the residue theorem: the main thing we have to be careful with is the exact argument of the function values. For $f_+$ we see that the argument of $z$ at the positive imaginary axis is $\pi/2$, and for the negative imaginary axis it is $3\pi/2$, while for $f_-$ the argument of $-z$ (not $z$) at the positive real axis is $-\pi/2$ and at the negative real axis it is $-3\pi/2$.
Now we can compute the residues at the simple poles $4i$ and $-2i$, and we find
$$\operatorname{Res}(f_+, 4i) = \frac{(4i)^{1/2}}{6i} = \frac{2e^{\pi i/4}}{6i}$$
$$\operatorname{Res}(f_-, 4i) = \frac{(-4i)^{1/2}}{6i} = \frac{2e^{-\pi i/4}}{6i}$$
$$\operatorname{Res}(f_+, -2i) = \frac{(-2i)^{1/2}}{-6i} = \frac{\sqrt2 e^{3\pi i/4}}{-6i}$$
$$\operatorname{Res}(f_-, -2i) = \frac{(2i)^{1/2}}{-6i} = \frac{\sqrt2 e^{-3\pi i/4}}{-6i}$$
$2\pi i$ times their sum is the sum of the contour integrals, so half that quantity is your integral:
$$\int_{-\infty}^\infty\frac{|x|^{1/2}}{(x - 4i)(x + 2i)} = \pi\left(\frac23\cos\frac\pi4 - \frac{\sqrt2}3\cos\frac{3\pi}4\right) = \pi\frac{\sqrt2 + 1}3.$$