(a) m₁ = f(0,0) = 0, m₂ = f(.1, .1 m₁) = f(.1,0) = -.1 so y₁ = y₀ + h (m₁ + m₂)/2 = -.005.

(b) Improved Euler has p=2, so the Richardson approximation is y_R = (4 y_.05 - y_.1)/(4-1) = -1.193590986. Taking this as exact, the error at h = 0.05 is .000230188. The error being approximately proportional to h², it will be about .00001 when h = .05 (.00001/.000230188)^1/2 = .010421454. You could round this off to .01.

(c) The overflow is caused by the fact that with this initial value, y goes to $\infty$ at some value of x between 0 and 2.