(2)

The relative velocity of the bus in the same direction of motion of scooty = $V_B-V_S=V_B-60$

The relative velocity of the bus in the opposite direction of motion of scooty = $V_B+V_S=V_B+60$

Now, distance travelled by bus in time (T) = $V_{B}T$

Now, setting up equations for the distances between the buses in both cases and solve for T and $V_B$.

$\frac{V_{B}T}{V_B-60}=30$                           ...(i)

$\frac{V_{B}T}{V_B+60}=10$                           ...(ii)

By solving equations (i) and (ii), we get

     $T=15\text{ min} \text{and} V_B=120\text{ km/h}$

(2)

Let $v_1$ is the velocity of Preeti on stationary escalator and $d$ is the distance travelled by her   $\therefore$  $v_1=\frac{d}{t_1}$

Again, let  $v_2$ be the velocity of escalator     $\therefore$   $v_2=\frac{d}{t_2}$

$\therefore$   Net velocity of Preeti on moving escalator with respect to the ground

           $v=v_1+v_2=\frac{d}{t_1}+\frac{d}{t_2}=d\left(\frac{{\displaystyle t_1+t_2}}{{\displaystyle t_1{t}_2}}\right)$

The time taken by her to walk up on the moving escalator will be 

$t=\frac{d}{v}=\frac{d}{d\left(\frac{t_1+t_2}{t_1{t}_2}\right)}=\frac{t_1{t}_2}{t_1+t_2}$