It equals $\Re \int_{-\infty}^\infty \frac{e^{iz}}{1-z^4}dz.$ The integrand $f(z)=\frac{e^{iz}}{1-z^4}$ has one simple pole$(z=i)$ in the upper half-plane, two simple poles$(z=\pm1)$ on the real axis.
$Res(f,i)=\frac{ze^{iz}}{-4} |_{z=i}=-\frac{ie^{-1}}{4}, Res(f,1)=-\frac{e^i}{4}, Res(f,-1)=\frac{e^{-i}}{4}.$
By contour integration, $\Re \int_{-\infty}^\infty \frac{e^{iz}}{1-z^4}dz = \Re[2\pi iRes(f,i)+\pi i\big(Res(f,1)+Res(f,-1)\big)]=\frac \pi 2(\sin 1+e^{-1}).$
But the answer gives "$\frac \pi 4(e^{-1}-\sin1)$". Where is wrong?
Let's note that due to the poles at $\pm 1$, the integral doesn't exist as a Lebesgue or improper Riemann integral. If we interpret it as a principal value integral, we do indeed obtain
$$\operatorname{v.p.} \int_{-\infty}^{+\infty} \frac{e^{iz}}{1-z^4}\,dz = 2\pi i \operatorname{Res}(f,i) + \pi i\bigl(\operatorname{Res}(f,1) + \operatorname{Res}(f,-1)\bigr) = \frac{\pi}{2e} + \frac{\pi}{2}\sin 1,$$
as you computed.
The mistake is in the provided answer. Without seeing what they did to arrive at that, I cannot tell what mistake was made.