Problem. 1
Problem. 2
Problem. 3
By hypothesis, For any reals $x$ and $t$, We are given $| f(t) - f(x) | \leq (t-x)^2$. Clearly:
$$ \begin{aligned} \frac{\displaystyle{|f(t) - f(x)|}}{\displaystyle{|t-x|}} \leq& \frac{\displaystyle{(t-x)^2}}{\displaystyle{|t-x|}} \\ \left | \frac{\displaystyle{f(t) - f(x)}}{\displaystyle{t-x}} \right | \leq& |t- x| \end{aligned} $$But $\lim_{t \rightarrow x} |t-x| = 0$ which concludes $\lim_{t \rightarrow x} \left | \frac{\displaystyle{f(t)-f(x)}}{\displaystyle{t-x}} \right | \leq 0$. Since the absolute value is always equal or greater than 0, We get also $\lim_{t \rightarrow x} \left | \frac{\displaystyle{f(t)-f(x)}}{\displaystyle{t- x}} \right | \geq 0$. Therefore $f'(x) = 0$ for any real $x$.