Answer
a) See the answer below.
b) $v_x =\sqrt{ L\left[\dfrac{2F_0 }{m}-\mu_0g \right] }$
Work Step by Step
a)
We can use the chain rule to find the acceleration of the block (to prove the given formula).
We know that
$$a_x=\dfrac{dv_x}{dt}$$
Multiplying the right side by $dx/dx$;
$$a_x=\dfrac{dv_x}{dt}\dfrac{dx}{dx}=\dfrac{dv_x}{dx}\dfrac{dx}{dt}$$
whereas $\dfrac{dx}{dt}=v_x$.
So that
$$\boxed{a_x=v_x\dfrac{dv_x}{dx}}\tag 1$$
b) Now we need to find the velocity expression as the block reaches $x=L$ from $x=0$.
We know that the pushing force $F_0$ is constant but the kinetic friction force is not.
$$\sum F_x=F_0-f_k=ma_x$$
$$F_0-\mu_kF_n=ma_x\tag 2$$
The net force exerted on the block in the vertical direction is zero.
$$\sum F_y=F_n-mg=0$$
$$F_n=mg$$
Plugging into (2);
$$F_0-\mu_kmg=ma_x $$
We know that $\mu_k=\mu_0\left(1-\dfrac{x}{L}\right)$
$$F_0-\mu_0\left(1-\dfrac{x}{L}\right)mg=ma_x $$
Solving for $a_x$;
$$a_x=\dfrac{F_0}{m}-\mu_0g\left(1-\dfrac{x}{L}\right) \tag 3 $$
Now we need to integrate (1) to $dx$;
$$\int a_x dx=\int \left(v_x\dfrac{dv_x}{dx}\right)dx=\int v_xdv_{x}$$
Plugging from (3);
$$\int_0^x \left[\dfrac{F_0}{m}-\mu_0g\left(1-\dfrac{x}{L}\right) \right]dx =\int_0^{v_x} v_xdv_{x}=\dfrac{v_x^2}{2}$$
Thus,
$$ \left[\dfrac{F_0x}{m}-\mu_0g\left(x-\dfrac{x^2}{2L}\right) \right] = \dfrac{v_x^2}{2}$$
$$ v_x^2= 2\left[\dfrac{F_0x}{m}-\mu_0g\left(x-\dfrac{x^2}{2L}\right) \right] $$
$$ v_x =\sqrt{ 2\left[\dfrac{F_0x}{m}-\mu_0g\left(x-\dfrac{x^2}{2L}\right) \right] }$$
Thus, the velocity at $x=L$ is
$$ v_x =\sqrt{ 2\left[\dfrac{F_0L}{m}-\mu_0g\left(L-\dfrac{L^2}{2L}\right) \right] }$$
$$ v_x =\sqrt{ 2\left[\dfrac{F_0L}{m}-\mu_0g\left(L-\dfrac{L }{2 }\right) \right] }$$
$$ v_x =\sqrt{ 2\left[\dfrac{F_0L}{m}-\mu_0g \dfrac{L }{2 } \right] }$$
$$ \boxed{v_x =\sqrt{ L\left[\dfrac{2F_0 }{m}-\mu_0g \right] }}$$
