Answer
$v_{max}=\sqrt{\frac{gr(v_0^2-\mu_s gr)}{gr+\mu_s v_0^2}}$
$v_{min}=\sqrt{\frac{gr(v_0^2+\mu_s gr)}{gr-\mu_s v_0^2}}$
Work Step by Step
When no friction is required, $F_N\sin(\theta)=m\frac{v_0^2}{r}$
$v_0^2=\frac{F_N\sin(\theta)r}{m}$
Also, $F_N\cos(\theta)=mg$
When there is friction, for maximum velocity, the force of friction will act in the direction down the slope. The forces acting on the car in the vertical direction are $F_N\cos(\theta)$, vertical component of the normal force, and $F_G$, the force of gravity.
$\sum F_y=F_N\cos(\theta)-mg-\mu F_N\sin(\theta)=0$
$F_N\cos(\theta)-\mu F_N\sin(\theta)=mg$
$mg-\mu m\frac{v_0^2}{r}=mg$
In the horizontal direction, the forces acting on the car are $F_f\cos(\theta)=\mu_s F_N\cos(\theta)$ and $F_N\sin(\theta)$
$\sum F_x=F_N\sin(\theta)-\mu_s F_N\cos(\theta)=\frac{mv_{max}^2}{r}$
$v_{max}^2=\frac{F_N\sin(\theta)r}{m}-\frac{\mu_s F_N\cos(\theta)r}{m}=v_0^2-\mu gr$
$\frac{gr}{v_{max}^2}=\frac{F_N\cos(\theta)-\mu F_N\sin(\theta)}{F_N\sin(\theta)-\mu_s F_N\cos(\theta)}=\frac{mg+\mu_s m\frac{v_0^2}{r}}{m\frac{v_0^2}{r}-\mu_s mg}$
$\frac{gr}{v_{max}^2}=\frac{gr+\mu_s v_0^2}{v_0^2-\mu_s gr}$
$v_{max}=\sqrt{\frac{gr(v_0^2-\mu_s gr)}{gr+\mu_s v_0^2}}$
For minimum velocity, the force of friction will act in the direction up the slope. Therefore,
$v_{min}=\sqrt{\frac{gr(v_0^2+\mu_s gr)}{gr-\mu_s v_0^2}}$