"For the first question, we wrote out the 3 times tables, (3,6,9, etc.)until the ninth multiple, because the missing numbers that we had to times it by 3 cannot be a two digit number. This is because if it is, there will be an unnecessary carried number, and you can't go on forever with carried numbers, since you would eventually get to the end, and there will still be a left out number.
First, we had to multiply 3 by e, that in 1, and the only multiple of 3 up to the ninth multiple that ends in 1, is 21 (3x7), so e is 7, and then we carried the 2 to d. dx3+2=7, something add 2 is 7, so it must end in 5. 3 times 5 equals 15, so d is 5. Now we carry the 1 to c, and 3xc has to end in a 4, because 4 add 1 equals 5, which is d. The only one that ends in 4 is 8x3 (24). Therefore, c is 8. Next, we carry the 2 to b. So that means, bx3+2= 8, and the only multiple of 3 now is 2 (6), 6 add 2 equals 8. This means you don't have to carry anything! Also that means b is 2. To work out (a) you have to do 3xa= a number that ends in 2. 4 times 3! Therefore, a is 4, and now you carry the 1, to 1. 3 times 1 add 1 equals 4!