1. Evaluate the integral $$\int_0^1 \int_y^1 dx \, dy$$. The inner integral with respect to $x$ is $$\int_y^1 dx = 1 - y$$. Now the outer integral becomes $$\int_0^1 (1 - y) dy = \left[y - \frac{y^2}{2}\right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2}$$. 2. Evaluate $$\int_0^1 \int_{-x}^3 y^{x-2} dy \, dx$$. For the inner integral over $y$, since $x$ is fixed, integrate $$\int_{-x}^3 y^{x-2} dy = \left[ \frac{y^{x-1}}{x-1} \right]_{-x}^3 = \frac{3^{x-1} - (-x)^{x-1}}{x-1}$$, assuming $x \neq 1$. We integrate this with respect to $x$ from 0 to 1: $$\int_0^1 \frac{3^{x-1} - (-x)^{x-1}}{x-1} dx$$. This integral is complicated; interpret $(-x)^{x-1}$ as for $x$ in (0,1), this is complex for negative base with non-integer exponent. With the problem focus, we keep the exact integral form or suggest numerical methods for evaluation. 3. Evaluate $$\int_0^{\pi/4} \int_0^{\sin x} dy \, dx$$. The inner integral $$\int_0^{\sin x} dy = \sin x$$. Outer integral: $$\int_0^{\pi/4} \sin x \; dx = \left[-\cos x \right]_0^{\pi/4} = -\cos(\pi/4) + \cos(0) = 1 - \frac{\sqrt{2}}{2}$$. 4. Evaluate $$\int_{-1}^1 \int_{y^2}^1 dx \, dy$$. Inner integral: $$\int_{y^2}^1 dx = 1 - y^2$$. Outer integral becomes: $$\int_{-1}^1 (1 - y^2) dy = \left[y - \frac{y^3}{3}\right]_{-1}^1 = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$$. Final Answers: 9. $$\frac{1}{2}$$ 10. Integral expression $$\int_0^1 \frac{3^{x-1} - (-x)^{x-1}}{x-1} dx$$ (numerical approximation may be needed). 11. $$1 - \frac{\sqrt{2}}{2}$$ 12. $$\frac{4}{3}$$