Q.If friction is zero, can a vehicle move on the road? Why are we not considering the friction in deriving the expression for the banking angle?

Banking a curved road means tilting its surface at some angle $\theta$ to the horizontal, so that the horizontal component of the (now-tilted) normal reaction, $N\sin\theta$, can itself supply some or all of the centripetal force, reducing the vehicle's dependence on the unreliable and limited force of friction. Ignoring friction, the relation $N\cos\theta=mg$ (vertical balance) combined with $N\sin\theta=\frac{mv^2}{r}$ (the horizontal, centripetal component) gives the single most important banking relation, $\tan\theta=\dfrac{v^2}{rg}$ -- usable either to find the one 'most safe speed' $v_s=\sqrt{rg\tan\theta}$ for a given road, or to find the required banking angle for a road designed for a given speed and radius.

In practice, friction (up to its limit) extends the SAFE range of speeds around $v_s$: at speeds below $v_s$, friction acts up the slope to prevent inward sliding, giving a lower speed limit $v_{min}=\sqrt{rg\frac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}}$; at speeds above $v_s$, friction acts down the slope to prevent outward skidding, giving an upper speed limit $v_{max}=\sqrt{rg\frac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}}$. A sufficiently steeply banked road ($\theta\geq45^\circ$) effectively has no finite upper speed limit at all, since the maximum possible $\mu_s$ (which is 1) is then always enough to prevent outward skidding at any speed.