It it was differentiable at $0$, then it would be true that:
$$ \lim_{x \to 0+} \frac{f(x)}{x} = \lim_{x \to 0-} \frac{f(x)}{x} = f'(0).$$
Above, the $x\to 0+$ means that we take the limit over the positive values of $x$, and likewise for $x \to 0-$.
Now, these two limits are easy to compute, and unfortunately are different. First, you have:
$$ \lim_{x \to 0-} \frac{f(x)}{x} = 0.$$
Secondly, you get with a little more work:
$$ \lim_{x \to 0+} \frac{f(x)}{x} = \lim_{x \to 0+} \frac{1}{1+x} = 1.$$
Now, these two values obviously can't be both equal to $f'(x)$. Thus, $f$ is not differentiable at $0$.
As for the latter part: NO, the derivative does not automatically have to be continuous. There is a Wikipedia article that you will surely find relevant. For example, the function $f$ given below is differentiable, but the derivative $f'$ is not continuous at $0$:
$$ f(x) = \begin{cases}
x^2 \sin \frac{1}{x} \quad& x > 0\\
0 \quad& x \leq 0
\end{cases}$$
The trick is that the term $x^2$ assures that $f$ goes to $0$ fast enough to have derivative $0$ at $0$, but the term $\sin \frac{1}{x}$ assures that close to $0$ the function has a large slope.
On the other hand, the derivative always has the mean value property, which is known as Darboux Theorem.
Ok, let me start pointing out (although you probably know that) that the classical mean value theorem
$$\frac{f(b) - f(a)}{b-a} = f'(\xi) \text{ for some } \xi \in (a,b)$$
does not hold in the Banach setting; the classical counterexample is $[0,1] \rightarrow \Bbb{C}, f(x) = e^{2\pi i x}$.
This is basically because (the proof of) Rolle's theorem uses the ordering on $\Bbb{R}$ in an essential way.
For the other question, there is one simple solution that does involve the supremum of the derivative rather than the integral. We have
$$\Vert f(b) - f(a) \Vert \leq (b - a) \cdot \sup_{\xi \in (a,b)} \Vert f'(\xi) \Vert.$$
This is proved by choosing (using Hahn-Banach) a bounded functional $\phi \in B'$ with $\Vert \phi \Vert \leq 1$ and $\phi(f(b) - f(a)) = \Vert f(b) - f(a) \Vert$ and then applying the classical mean-value theorem to $\phi \circ f$.
For the question concerning the integral, let us assume that $\int_a^b \Vert f'(t) \Vert dt < \infty$ (otherwise, the estimate is trivial). Note that $f' : (a,b) \rightarrow B$ is measurable as it is the pointwise limit of the measurable functions $x \mapsto n \cdot (f(x + \frac{1}{n}) - f(x)) \cdot \chi_{(a, b-\frac{1}{n})}$. I only added the indicator function to avoid the problem $x + \frac{1}{n} \notin (a,b)$ although $x \in (a,b)$.
Let us again take $\phi \in B'$ arbitrary. Then the hypothesis also holds for $f_\phi := \phi \circ f$ and we also have $(\phi \circ f)' = \phi \circ f' \in L^1((a,b))$.
Now Theorem 7.21 in Rudin, Real and Complex Analysis implies
$$\phi(f(b) - f(a)) = f_\phi(b) - f_\phi(a) = \int_a^b f_\phi'(x) dx = \phi\left(\int_a^b f'(x) dx \right).$$
Actually, Rudin requires $f_\phi$ to be differentiable on $[a,b$] instead of $(a,b)$, but then the above applies to $a+\frac{1}{n}$ and $b-\frac{1}{n}$ instead of $a,b$ and we can take the limit as $f$ is continuous and $f'$ is integrable.
As $\phi \in B'$ was arbitrary (a corollary of) Hahn Banach implies
$$f(b) - f(a) = \int_a^b f'(x) dx.$$
Taking norms, you arrive at the desired conclusion.
In short, we only needed to assume (Lebesgue) integrability of the derivative, not $f \in C^1$. Note that this is always satisfied if $f'$ is bounded on $(a,b)$.
Best Answer
I’ll assume you want to ask the function to be continuous at $0$; otherwise, the example by @player3236 gives a counterexample because the lack of continuity at $0$ messes you up even before you get started.
Let $f(x)$ be defined by $$f(x) = \left\{\begin{array}{ll} |x|\sin\left(\frac{1}{x}\right) &\text{if }x\neq 0,\\ 0&\text{if }x=0. \end{array}\right.$$ Then for any $x\neq 0$ we have $$f(-x) = |-x|\sin\left(\frac{1}{-x}\right) = -|x|\sin\left(\frac{1}{x}\right) = -f(x),$$ so $f$ is odd. It is also differentiable everywhere other than $0$, and is continuous at $0$. However, $$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x} = \lim_{x\to 0^+}\sin\left(\frac{1}{x}\right)$$ does not exist, so it does not have a derivative at $x=0$.