Hello, please consider the following.
PART 2
Let's note [tex](x_i)_{1\leq i\leq 11}[/tex] the 11 numbers.
We can write the following
[tex]x_1^2+...+x_{11}^2=(x_1+1)^2+...+(x_{11}+1)^2\\\\=x_1^2+...+x_{11}^2+2(x_1+...+x_{11})+11\\\\\text{So } 2(x_1+...+x_{11})+11 = 0\\\\(x_1+2)^2+...+(x_{11}+2)^2=x_1^2+...+x_{11}^2+4(x_1+...+x_{11})+4*11\\\\\text{As we know that }2(x_1+...+x_{11})+11 = 0\\4(x_1+...+x_{11})+4*11=-11*2+4*11=22[/tex]
So the sum of squares changes by
[tex]\boxed{ \ 22 \ }[/tex]
PART 4
Let's say that we have n points at the beginning.
We will add n-1 points the first time, we will get n + n - 1 = 2n -1 points.
And then, the second time we add 2n - 1 - 1 points and we get
2n - 1 +2n - 2 = 4n - 3.
Finally, we do it a last time, we add 4n - 3 - 1 points and we get 4n - 3 + 4n - 4 points = 8n - 7 and it must be 65.
So, 8n -7 = 65 <=> 8n = 65+7=72 <=> n = 72/8=9
[tex]\boxed{= \ 9}[/tex]
Thank you