Suppose we take the centre number as "x" This will give us a variable to compare all the numbers.
To make it easier, we will take the addition of the 4 corners as "y". The top left corner will always be x - 8, and the top right would be x-6. The bottom left would be
x + 6, and the bottom right would be x + 8. So if you add them all together, you would get {(x -8) +(x - 6) + (x + 6) + (x + 8)}. When you open the brackets and group the numbers and the variables, you will get 4x + 0 = y. Therefore, the four corners will always be 4 times the middle number.
Part 2.
Using the same variable i.e middle number being "x". Now in terms of "x", other numbers will be :
1. Top left : x-8
2. Top right : x-6
3. Bottom left : x+6
4. Bottom right : x+8
5. Top middle : x-7
6. Bottom middle : x+7
7. Middle left : x-1
8. Middle right : x+1
Now adding all these nos will give us an equation :
X-8+x-6+x+8+x+6+x-7+x+7+x-1+x+1 = 8x
Hence the number will always be 8 times the centre number.
Part 3.
Again using the same method and adding all numbers. We will come to 6x and hence in this scenario, the sum will always be 6 times the middle number.
Part 4.
When you look at the columns in a calender, you will see that the left column consists of the lowest number, the middle column will have the numbers in between, and the right column will have the highest number. Hence, No matter which numbers you will choose, the numbers will always one low number, one middle number, and one high number. You will then notice that these numbers always cut each other, and you will invariably be left with 3x. For the given example, they have chosen the low number as (x - 8) ,the high number as (x + 1) and the middle number as (x + 7). The top left, the middle right and middle bottom. If you do the maths, you will be left with 3x + 0. Therefore, the 3 numbers will always equal to 3 times the middle number.