The 4th answer is pretty obvious on putting values, we get n = 44. Hence, largest prime factor=11. The 3rd answer is also pretty easy....

Let the polynomial be a + bx + cx^2 + dx^3 + .......

Now, since the polynomial is divisible by n for all integral values, it can be proved that a must be 0. Now, the polynomial is bx + cx^2 + dx^3 + ......

Plug in 0, we get 00 as the answer.

In the 5th sum, let the number be 10a + b. Then, the number formed by interchanging the digits will be 10b+a. Now given, 10a + b + 10b + a = 11(a+b) is a perfect square, say N^2

=> N^2 is a multiple of 11. Also, a+b is less than 19 since both are digits. hence , 11(a+b) is less than 209. This condition is only satisfied when N = 11 and N^2 = 121.

=> a+b = 11, The ways for which we can solve this is 4!/3! * 2! ways, that is 8 ways.

Sum number 6, as mentioned in early comments, can be...