Solution:-

(i) We have to find the ratios AD/AB, DE/BC,

From the question it is given that, AD/DB = 3/2

Then, DB/AD = 2/3

Now add 1 for both LHS and RHS we get,

(DB/AD) + 1 = (2/3) + 1

(DB + AD)/AD = (2 + 3)/3

From the figure (DB + AD) = AB

So, AB/AD = 5/3

Now, consider the ∆ADE and ∆ABC,

∠ADE = ∠B … [corresponding angles are equal]

∠AED = ∠C … [corresponding angles are equal]

Therefore, ∆ADE ~ ∆ABC … [by AA similarity]

Then, AD/AB = DE/BC = 3/5

(ii) Now consider the ∆DEF and ∆CBF

∠EDF = ∠BCF … [because alternate angles are equal]

∠DEF = ∠FBC … [because alternate angles are equal]

∠DFE = ∠ABFC … [because vertically opposite angles are equal]

Therefore, ∆DEF ~ ∆CBF

So, EF/FB = DE/BC = 3/5

(iii) we have to find the ratio of the areas of ∆DEF and ∆CBF,

We know that, Area of ∆DFE/Area of ∆BFC = DE²/BC²

Area of ∆DFE/Area of ∆BFC = (DE/BC)²

Area of ∆DFE/Area of ∆BFC = (3/5)²

Area of ∆DFE/Area of ∆BFC = 9/25

Therefore, the ratio of the areas of ∆DEF and ∆CBF is 9: 25.