Differentiable function and tangent line 
a. Prove that differentiable function $f(x)$ on $[a,b]$ and its tangent line $T(x)$ at $a$ satisfies $|f(x) - T(x)|\le C|x-a|$ where $C=\sup\limits_{a\le y \le x}|f'(y) -f'(a)|$.

No idea! I am guessing using mean value theorem: $$\frac{f(x)-f(a)}{x-a}=f'(c)$$

b. If $f$ is $C^2$, refine this statement to: $$|f(x) -T(x)| \leq D|x-a|^2$$ where $D = \sup\limits_{a\le y \le x}|f''(y)|$.

Some useful facts:
$T(x) = f(x_0)+f'(x_0)\cdot(x-x_0)$
 A: *

*We apply the Taylor-Lagrange formula to $f$ with the first  order : there's $\alpha\in(a,x)$ such that
$$f(x)=f(a)+(x-a)f'(\alpha)$$
so we find the first result
$$|f(x)-T(x)|=(x-a)|f'(\alpha)-f'(a)|\leq(x-a)\sup_{a\leq y\leq x}|f'(y)-f'(a)|$$

*We apply the Taylor-Lagrange formula to $f$ with the second order: there's $\alpha\in(a,x)$ such that
$$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(\alpha)=T(x)+\frac{(x-a)^2}{2!}f''(\alpha),$$
so we find the second result
$$|f(x)-T(x)|=\frac{(x-a)^2}{2!}|f''(\alpha)|\leq\frac{(x-a)^2}{2!}\sup_{a\leq y\leq x}|f''(y)|. $$

A: Introduce the function 
$$
g(x):=f(x)-T(x)=f(x)-f(a)-(x-a)f'(a).
$$
Then note it has the same differentiability properties as $f$, with $g(a)=0$, $g'(x)=f'(x)-f'(a)$, $g'(a)=0$ , and $g''(x)=f''(x)$.
Also note that we only need $f$ twice differentiable on $(a,b)$ to get b.
a. Apply the mean value theorem to $g$ on $[a,x]$. This gives $y_x\in(a,x)$ such that $$g(x)=g(x)-g(a)=g'(y_x)(x-a).$$ The result follows.
b. Apply the mean value theorem to $g'$ on $[a,y_x]$. T yields $z_x$ in $(a,y_x)\subseteq[a,x]$ such that $$g'(y_x)=g'(y_x)-g'(a)=g''(z_x)(x-a).$$ Hence $g(x)=g''(z_x)(x-a)^2$ and the result follows.
A: We have $f(x)=\int_a^x f'(t)dt+f(a)$ and $T(x)=f'(a)(x-a)+f(a)$. Thus
$$\begin{align}
\frac{|f(x)-T(x)|}{|x-a|} &=\frac{\left|\int_a^xf'(t)dt-f'(a)(x-a)\right|}{|x-a|}\\
&=\left|\int_a^x\frac{f'(t)-f'(a)}{x-a}dt\right|\\
&\leq \int_a^x\left|\frac{f'(t)-f'(a)}{x-a}\right|dt\\
&\leq \int_a^x\frac{C}{x-a}dt=C
\end{align}$$
so $|f(x)-T(x)|\leq C|x-a|$.
Edit: As julien points out, this only works for continuously differentiable functions. His answer is correct in general.
