Usaremos la fórmula: $K = \frac{|y''|}{(1 + (y')^2)^{3/2}}$.

Inciso (a): $y = \frac{x^3}{3} \implies y' = x^2, y'' = 2x$.

• Para $x = 0$: $y'=0, y''=0 \implies K = 0$.
• Para $x = 1$: $y'=1, y''=2 \implies K = \frac{2}{(1+1)^{3/2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
• Para $x = -2$: $y'=4, y''=-4 \implies K = \frac{|-4|}{(1+16)^{3/2}} = \frac{4}{17\sqrt{17}} = \frac{4\sqrt{17}}{289}$.

Inciso (b): $y = \frac{x^2}{4a} \implies y' = \frac{x}{2a}, y'' = \frac{1}{2a}$.

• Para $x = 0$: $y'=0, y''=\frac{1}{2a} \implies K = \frac{1/2a}{(1+0)^{3/2}} = \frac{1}{2a}$.
• Para $x = 2a$: $y'=1, y''=\frac{1}{2a} \implies K = \frac{1/2a}{(1+1)^{3/2}} = \frac{1/2a}{2\sqrt{2}} = \frac{1}{4a\sqrt{2}} = \frac{\sqrt{2}}{8a}$.

Inciso (c): $y = \sin x \implies y' = \cos x, y'' = -\sin x$.

• Para $x = 0$: $y'=1, y''=0 \implies K = 0$.
• Para $x = \pi/2$: $y'=0, y''=-1 \implies K = \frac{|-1|}{(1+0)^{3/2}} = 1$. (Nota: La curvatura es magnitud, el signo depende de la orientación).

Inciso (d): $y = e^{-x^2} \implies y' = -2xe^{-x^2}, y'' = e^{-x^2}(4x^2 - 2)$.

• Para $x = 0$: $y'=0, y'' = 1(0-2) = -2 \implies K = \frac{|-2|}{(1+0)^{3/2}} = 2$.

$$ \boxed{\text{Resultados: (a) } 0, \frac{\sqrt{2}}{2}, \frac{4\sqrt{17}}{289}; \text{ (b) } \frac{1}{2a}, \frac{\sqrt{2}}{8a}; \text{ (c) } 0, 1; \text{ (d) } 2} $$