Length of the train = 40 m = 0.04 km
Length of the bridge = 200 m = 0.2 km
Speed of the train = 36 km/hr

The total time taken by the train to completely pass over the bridge is given by time for the front of the train to travel 200 m at 36 km/hr + time for the back of train to travel 40 m at a speed of 36 km/hr. 

Therefore, 

Total time taken = $\frac{{\mathrm{s}}_1}{\mathrm{v}}+\frac{s_2}{v}$ 

                         $$\begin{gathered} = \frac{1}{v}\left[s_1 + s_2\right] \\ = \frac{0.2}{36}+\frac{0.04}{36} \\ = 0.005 + 0.001 \\ = 0.006 hrs \\ = 24 sec \end{gathered}$$

Let, the total distance covered the car be 'd'. 
Given, 
Speed in the first half = 40 km/hr
Speed in the second half of distance = 60 km/hr

Let the time taken by the car in the first half = t₁
Therefore, 
Time, ${\mathrm{t}}_1 = \frac{\mathrm{d}/2}{40}= \frac{\mathrm{d}}{80}$

Time taken by car in the second half, t₂ = $\frac{{\displaystyle \raisebox{1ex}{$\mathrm{d}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{60}=\frac{d}{120}$

Now, average speed = $\frac{\mathrm{Total} \mathrm{distance} \mathrm{travelled}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}$

                              $$\begin{gathered} = \frac{\mathrm{d}}{{\displaystyle \frac{\mathrm{d}}{80}}+{\displaystyle \frac{\mathrm{d}}{120}}} \\ = \frac{80\times 120}{80+120} \\ = \frac{9600}{200} \\ = 48 \mathrm{km}/\mathrm{hr} \end{gathered}$$ 

Let the total distance travelled to and fro be d
Speed while ging from town A to town B = 40 km/hr
Speed while coming back from town B to town A = 60 km/hr
Time taken to go to town B ,t₁ = $\frac{Dis\tan ce}{Speed}$

                                           t₁ = $\frac{{\displaystyle \raisebox{1ex}{$\mathrm{d}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{40} = \frac{d}{80}$

Time taken to return to town A, t₂ = $\frac{{\displaystyle \raisebox{1ex}{$\mathrm{d}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{60}=\frac{d}{120}$

Average speed = $\frac{\mathrm{Total} \mathrm{distance}}{\mathrm{toatl} \mathrm{time}}$ 

                      $$\begin{gathered} =\frac{\mathrm{d}}{{\displaystyle \frac{\mathrm{d}}{80}}+{\displaystyle \frac{\mathrm{d}}{120}}} \\ = \frac{80\times 120}{80+120} \\ =\frac{9600}{200} \\ = 48 km/hr \end{gathered}$$

Given, 
Velocity = 72 km/hr

So, 
Velocity in m/s will be given by, 

              v = $72 \times \frac{1000}{3600} = 20 \mathrm{m}/\mathrm{s}$

Given, 
Distance travelled by the car = 270 km 

Total time for travel = 4.5 hours

i) Speed of the car, v  = $\frac{\mathrm{Distance} \mathrm{travelled}}{\mathrm{Time} \mathrm{taken}}$ 

                                =$\frac{270}{4.5} = 60 km/hr$


ii) Speed = 60 km/hr

Time taken = 7 hours

Distance travelled = $\mathrm{Speed} \times \mathrm{time}$ 

                           =  420 km

(iii) Speed = 60 km/hr

Distance to be travelled = 300 km

Time taken = $\frac{\mathrm{Distance}}{\mathrm{Speed}}$

                  = $\frac{300}{60}= 5 hours$