1. **Problem (a):** Evaluate the integral $$\int_0^{2\pi} \frac{d\theta}{\frac{5}{4} + \sin \theta}.$$\n\n2. **Formula and rules:** For integrals of the form $$\int_0^{2\pi} \frac{d\theta}{a + b \sin \theta}$$ with $$|a| > |b|,$$ the result is $$\frac{2\pi}{\sqrt{a^2 - b^2}}.$$\n\n3. **Apply to our problem:** Here, $$a = \frac{5}{4} = 1.25$$ and $$b = 1.$$ Check $$|a| > |b|$$: $$1.25 > 1$$ is true, so formula applies.\n\n4. **Calculate denominator:** $$\sqrt{a^2 - b^2} = \sqrt{\left(\frac{5}{4}\right)^2 - 1^2} = \sqrt{\frac{25}{16} - 1} = \sqrt{\frac{25}{16} - \frac{16}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}.$$\n\n5. **Final answer for (a):** $$\int_0^{2\pi} \frac{d\theta}{\frac{5}{4} + \sin \theta} = \frac{2\pi}{\frac{3}{4}} = 2\pi \times \frac{4}{3} = \frac{8\pi}{3}.$$\n\n---\n\n6. **Problem (b):** Evaluate $$\int_C \frac{dz}{\sinh z}$$ where $$C := |z| = 2.$$\n\n7. **Key points:** The contour is the circle of radius 2 centered at 0. The function $$\frac{1}{\sinh z}$$ has singularities where $$\sinh z = 0,$$ i.e., at $$z = k\pi i$$ for integers $$k.$$\n\n8. **Singularities inside $$|z|=2$$:** The points $$0, \pi i, -\pi i$$ lie inside the circle since $$|0|=0 < 2,$$ $$|\pi i|=\pi \approx 3.14 > 2,$$ so only $$z=0$$ is inside the contour.\n\n9. **Residue at $$z=0$$:** Near zero, $$\sinh z \approx z,$$ so $$\frac{1}{\sinh z} \approx \frac{1}{z}.$$ The residue at $$z=0$$ is 1.\n\n10. **By the residue theorem:** $$\int_C \frac{dz}{\sinh z} = 2\pi i \times (\text{sum of residues inside } C) = 2\pi i \times 1 = 2\pi i.$$\n\n---\n\n11. **Problem (c):** Evaluate $$\int_C z \exp \frac{1}{z} dz$$ where $$C := |z| = 1.$$\n\n12. **Expand the integrand:** $$\exp \frac{1}{z} = \sum_{n=0}^\infty \frac{1}{n!} \frac{1}{z^n} = 1 + \frac{1}{z} + \frac{1}{2! z^2} + \frac{1}{3! z^3} + \cdots.$$\n\n13. **Multiply by $$z$$:** $$z \exp \frac{1}{z} = z \left(1 + \frac{1}{z} + \frac{1}{2! z^2} + \frac{1}{3! z^3} + \cdots \right) = z + 1 + \frac{1}{2! z} + \frac{1}{3! z^2} + \cdots.$$\n\n14. **Residue theorem:** The integral over $$C$$ equals $$2\pi i$$ times the coefficient of $$\frac{1}{z}$$ in the Laurent expansion. Here, the coefficient of $$\frac{1}{z}$$ is $$\frac{1}{2!} = \frac{1}{2}.$$\n\n15. **Final answer for (c):** $$\int_C z \exp \frac{1}{z} dz = 2\pi i \times \frac{1}{2} = \pi i.$$