1. **State the problem:** Evaluate the triple integral $$\int_0^\pi \int_0^{\frac{\pi}{2}} \int_0^1 \left(z \sin x + z \cos y\right) \, dz \, dy \, dx.$$\n\n2. **Separate the integral:** Because integration is linear, $$\int_0^1 (z \sin x + z \cos y) \, dz = \sin x \int_0^1 z \, dz + \cos y \int_0^1 z \, dz.$$\n\n3. **Evaluate the integral over $z$:** $$\int_0^1 z \, dz = \left.\frac{z^2}{2}\right|_0^1 = \frac{1}{2}.$$\n\n4. **Rewrite the integral:** $$\sin x \cdot \frac{1}{2} + \cos y \cdot \frac{1}{2} = \frac{1}{2}(\sin x + \cos y).$$\n\n5. **Substitute back into the integral over $y$ and $x$: ** $$\int_0^\pi \int_0^{\frac{\pi}{2}} \frac{1}{2}(\sin x + \cos y) \, dy \, dx = \frac{1}{2} \int_0^\pi \int_0^{\frac{\pi}{2}} (\sin x + \cos y) \, dy \, dx.$$\n\n6. **Split the integral over $y$: ** $$\frac{1}{2} \int_0^\pi \left( \int_0^{\frac{\pi}{2}} \sin x \, dy + \int_0^{\frac{\pi}{2}} \cos y \, dy \right) dx.$$\n\n7. **Evaluate inner integrals over $y$: ** - Since $\sin x$ does not depend on $y$, $$\int_0^{\frac{\pi}{2}} \sin x \, dy = \sin x \cdot \frac{\pi}{2}.$$\n - For the cosine term, $$\int_0^{\frac{\pi}{2}} \cos y \, dy = \left.\sin y\right|_0^{\frac{\pi}{2}} = 1 - 0 = 1.$$\n\n8. **Substitute these back:** $$\frac{1}{2} \int_0^\pi \left( \sin x \frac{\pi}{2} + 1 \right) dx = \frac{1}{2} \int_0^\pi \left( \frac{\pi}{2} \sin x + 1 \right) dx.$$\n\n9. **Split and evaluate integral over $x$: ** $$\frac{1}{2} \left( \frac{\pi}{2} \int_0^\pi \sin x \, dx + \int_0^\pi 1 \, dx \right).$$\n\n10. **Compute each integral: ** - $$\int_0^\pi \sin x \, dx = \left.-\cos x\right|_0^\pi = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2.$$\n - $$\int_0^\pi 1 \, dx = \pi.$$\n\n11. **Substitute back:** $$\frac{1}{2} \left( \frac{\pi}{2} \cdot 2 + \pi \right) = \frac{1}{2} (\pi + \pi) = \frac{1}{2} \cdot 2\pi = \pi.$$\n\n**Final answer:** $$\pi.$$