It appears that you are being asked to demonstrate that each step is correct.
For example, for (i) you could write:
$\displaystyle (1+x)^n=\sum_{k=0}^n{n \choose k}x^k={n \choose 2}x^2+\left(1+nx+\sum_{k=3}^n{n \choose k}x^k \right)$
For $k\ge0$ it must be true that:
$\displaystyle 1+nx+\sum_{k=3}^n{n \choose k}x^k\ge0$
and so by adding $\displaystyle {n \choose 2}x^2$ to both sides, we obtain:
$\displaystyle (1+x)^n\ge{n \choose 2}x^2=\frac{n(n-1)}{2}x^2$
For (ii) you could begin with:
$\displaystyle \frac{n}{2}\ge1$ for $n\ge2$
Now add $\displaystyle \frac{n}{2}$ to both sides:
$\displaystyle n\ge1+\frac{n}{2}$
Subtract through by 1:
$\displaystyle n-1\ge\frac{n}{2}$
Now, multiply through by $\displaystyle \frac{n}{2}$
$\displaystyle \frac{n(n-1)}{2}\ge\frac{n^2}{4}$
From (i) we have:
$\displaystyle (1+x)^n\ge\frac{n(n-1)}{2}x^2$
And so on the right, replacing $\displaystyle \frac{n(n-1)}{2}$ with $\displaystyle \frac{n^2}{4}$ we obtain:
$\displaystyle (1+x)^n\ge\frac{n^4}{4}x^2$
Multiplying through by $\displaystyle \frac{4}{n^2}$ there results:
$\displaystyle \frac{4}{n^2}(1+x)^n\ge x^2$
$\displaystyle x^2\le\frac{4}{n^2}(1+x)^n$ where $n\ge2,\,x\ge0$
Now for (iii), it is just a matter of applying the result of Problem 1(ii) (which you haven't provided).
I didnt know I could do that, that may help in future sums :) I was just making sure I was doing it in the right way, we get given so many definitions and corollaries etc :)