Your answers seem to be correct.
1. The limit as x→0 of (eˣ-1)/x is 1 because eˣ≈1+x when x is small so (1+x-1)/x=x/x=1, and g(0)=1, so there is continuity at x=0.
2. Using the integration factor e^(x³/3) and substituting the initial conditions, the integration constant is -1, leading to ye^(x³/3)=e^(x³/3)-1, which is y=1-e^-(x³/3).
3. The absolute convergence test aᵣ₊₁/aᵣ<0 and 0<1 so aᵣ₊₁/aᵣ<1.
4. When a₁=x+1, |a₂| etc < x+1, making the radius 1.