1. **State the problem:** We want to compute the second approximation $y_2$ for the initial value problem $$y' = e^{-t} + 5y^2$$ with initial condition $y(0) = 0.00$, step size $h=0.5$, and given $w_1 = 0.6520$. 2. **Recall the 2-Stage Adams-Bashforth method formula:** $$w_{n+1} = w_n + \frac{h}{2} \left(3f(t_n,w_n) - f(t_{n-1},w_{n-1})\right)$$ where $f(t,y) = e^{-t} + 5y^2$. 3. **Calculate $f(t_0,w_0)$:** Given $t_0=0$, $w_0=0.00$, $$f(t_0,w_0) = e^{0} + 5 \times 0^2 = 1 + 0 = 1.0000$$ 4. **Calculate $f(t_1,w_1)$:** Given $t_1 = t_0 + h = 0.5$, $w_1=0.6520$, $$f(t_1,w_1) = e^{-0.5} + 5 \times (0.6520)^2$$ Calculate each term: $$e^{-0.5} \approx 0.6065$$ $$(0.6520)^2 = 0.4251$$ $$5 \times 0.4251 = 2.1255$$ So, $$f(t_1,w_1) = 0.6065 + 2.1255 = 2.7320$$ 5. **Apply the Adams-Bashforth formula to find $w_2$:** $$w_2 = w_1 + \frac{0.5}{2} \left(3 \times 2.7320 - 1.0000\right)$$ Calculate inside the parentheses: $$3 \times 2.7320 = 8.1960$$ $$8.1960 - 1.0000 = 7.1960$$ Multiply by $\frac{h}{2} = 0.25$: $$0.25 \times 7.1960 = 1.7990$$ Add to $w_1$: $$w_2 = 0.6520 + 1.7990 = 2.4510$$ 6. **Final answer:** Rounded to 2 decimal places, $$\boxed{y_2 = 2.45}$$