You got stuck with second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis):
$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$
But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.
Substitute:
$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$
Set:
$$\frac km=\omega^2$$
$$\ddot{X}+\omega^2 X=0$$ This classic differential equation solves to:
$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:
Assume at $t=0$, $x=0 \implies X(0)=-F$
$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.
$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin(\phi)=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$
Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.
So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it.