1. The problem is to evaluate the contour integral $$\oint \frac{\sin z}{(z+i)^3} \, dz$$ around a closed contour enclosing the singularity at $z = -i$. 2. Notice that the integrand has a pole of order 3 at $z = -i$. 3. By the residue theorem, the integral equals $2\pi i$ times the residue of the function at $z = -i$. 4. The residue at a pole of order 3 is given by: $$\text{Res}_{z=-i} f(z) = \frac{1}{2!} \lim_{z \to -i} \frac{d^2}{dz^2} \left[(z+i)^3 \frac{\sin z}{(z+i)^3}\right] = \frac{1}{2} \lim_{z \to -i} \frac{d^2}{dz^2} \sin z$$ 5. Since $(z+i)^3$ cancels, we need the second derivative of $\sin z$ evaluated at $z = -i$: $$\frac{d}{dz} \sin z = \cos z$$ $$\frac{d^2}{dz^2} \sin z = -\sin z$$ 6. Evaluate at $z = -i$: $$-\sin(-i) = -(-\sin i) = \sin i$$ 7. Recall that $\sin i = i \sinh 1$. 8. Therefore, the residue is: $$\frac{1}{2} \sin i = \frac{1}{2} i \sinh 1$$ 9. Finally, the integral is: $$\oint \frac{\sin z}{(z+i)^3} \, dz = 2\pi i \times \frac{1}{2} i \sinh 1 = \pi i^2 \sinh 1 = -\pi \sinh 1$$ **Answer:** $$\boxed{-\pi \sinh 1}$$