It is cute how they say please everywhere. First note the triangles ADE and ABC are similar. So their corresponding sides have lengths in the same ratio.
(i) 10/(4+2) = DE/2 so DE = 10/3 (which is not equal to 15)
(ii) intuitively, the sides of the ABC are (4+2)/2 = 3 times longer than sides of ADE, so the area of ABC is 3^2=9 times bigger.
More precisely, draw the altitude from the vertex A to BC. Mark P the intersection of the altitude with DE and Q the intersection of the altitude with AB. By similarity of triangles APD and AQB you have AP/AQ = AD/AB = 1/3.
Now the area of ABC is equal to BC*AQ/2 = DE*3*AP*3/2 = (DE*AP/2)*9 = m*9.
(iii) why new symbol for something already denoted by m?
The area of DECB is the area of ABC minus the area of ADE. So use (ii) to get the result.