The answer to the first question is YES. In fact, one can show the following stronger statement :
THEOREM. There exists an increasing homeomorphism $f: [0,1] \to [0,1]$, such that $f'(q)=0$ for all $q\in {\mathbb Q}$ and any irrational $x\in [0,1]$ is contained in arbitrary small intervals $[u,v]$ with $\frac{f(v)-f(u)}{v-u} \geq 1$ (so that $f'(x)$, if defined, is $\geq 1$).
We use a familiar "inbreeding", piecewise and iterative construction, enumerating the rationals and imposing $f'(q)=0$ for each $q$ one by one, keeping at the same time a tight net of values $x,y$ such that $\frac{f(y)-f(x)}{y-x} \geq 1$. Formally :
DEFINITION. Let $f,g$ be two function $[a,b] \to {\mathbb R}$. We say that $(f,g)$ is a claw on $[a,b]$ when
(1) $f(a)=g(a),f(b)=g(b),f'(a)=g'(a)=f'(b)=g'(b)=0$ and
(2) $f(x)<g(x)$ when $a<x<b$.
If, in addition, one has
(3) For any $x\in ]a,b[$, there are $u<v$ in $]a,b[$ with $x\in [u,v]$, $\frac{f(v)-g(u)}{v-u} \geq 1$,
then we say that $(f,g)$ is a tight claw.
FUNDAMENTAL LEMMA. If $(f,g)$ is any claw on $[a,b]$, then there is another claw $(F,G)$ with $f \leq F \leq G \leq g$ on $[a,b]$ and $(F,G)$ is tight. We may further impose that the supremum norm $||F-G||_{\infty}$ be $\lt \epsilon$ for $\epsilon$ as small as we want.
{\bf Proof of theorem using fundamental lemma}. We construct by induction two sequences $f_n,g_n$ of homeomorphisms $[0,1] \to [0,1]$ such that :
(i) $f_n \leq g_n$ on $[0,1]$
(ii) $(f_n,g_n)$ is a tight claw on each interval $I_{n,k}=[\frac{k}{n!},\frac{k+1}{n!}]$ for $0 \leq k\ \leq n!$.
(iii) $||f_n-g_n||_{\infty} \leq \frac{1}{n}$.
We start with $f_0(x)=4x^3-3x^4$ and $g_0(x)=3x^2-2x^3$. Suppose now that $(f_n,g_n)$ have already been constructed. The two maps $f_n$ and $g_n$ coincide on
the finite set $X_n=\lbrace \frac{k}{n!} | 0 \leq k \leq n! \rbrace$. There is an increasing map $\phi : X_{n+1} \to [0,1]$, extending $f_{|X_n}=g_{|X_n}$, such that $f_n \lt \phi \lt g_n $ on $X_{n+1} \setminus X_n$.
On each interval $I_{n+1,k}$, there is a claw $(p_k,q_k)$ with $f_n \leq p_k \leq q_k \leq g_n$, and $p_k$ nad $q_k$ coincide with $\phi$ on the extremities of the interval. By fundamental lemma, there is a tight claw $(r_k,s_k)$ on $I_{n+1,k}$ with $p_k \leq r_k \leq s_k \leq q_k$, and $||r_k-s_k||_{\infty} \leq \frac{1}{n}$.
We may then take $f_{n+1}$ ($g_{n+1}$, respectively) to be the unique function that coincides with $r_k$ ($s_k$, respectively) on $I_{n+1,k}$. This completes the inductive construction.
Then by (i) and (iii) we have ${\sf sup}(f_n)={\sf inf}(g_n)$ ; let us call $f$ this function. Then $(f_n)$ nad $(g_n)$ converge uniformly towards $f$, and $f$ is a homeomorphism $[0,1] \to [0,1] $. We have for $x \in I_{n,k}$ and $q=\frac{k}{n!}$, and any $m\geq n$,
$$
f_m(q)=g_m(q)=f(q), \ \frac{f_n(x)-f(q)}{x-q} \leq \frac{f(x)-f(q)}{x-q} \leq \frac{g_n(x)-f(q)}{x-q}
$$
whence $f'(q)=0$ on the right. Similarly, $f'(q)=0$ on the left.
The fundamental lemma can be shown from the following :
LEMMA 1. Let $(f,g)$ be a claw on $[a,b]$. Then there are three increasing sequences $(x_k)_{k\in \mathbb Z},(y_{1,k})_{k \in \mathbb Z}$ and $(y_{2,k})_{k \in \mathbb Z}$, with
$$
{\sf lim}_{k \to -\infty} x_k=a, \
{\sf lim}_{k \to +\infty} x_k=b, \
$$
$$
{\sf lim}_{k \to -\infty} y_{1,k}={\sf lim}_{k \to -\infty} y_{2,k}=f(a)=g(a),\
{\sf lim}_{k \to +\infty} y_{1,k}={\sf lim}_{k \to +\infty} y_{2,k}=f(b)=g(b)
$$
and
$$
\frac{y_{1,k+2}-y_{2,k}}{x_{k+2}-x_k} \leq 1, f(x_k) \lt y_{1,k} \lt y_{2,k} <g(x_k), |y_{2,k}-y_{1,k}| \lt \epsilon
$$
{\bf Proof of fundamental lemma from lemma 1.} Set $F(x_k)=y_{1,k}, G(x_k)=y_{2,k}$ for all $k\in \mathbb Z$ and interpolate in-between (take $F$ and $G$ affine on $[x_k,x_{k+1}]$, for instance).
Finally, lemma 1 follows from the iteration of
LEMMA 2. Let $(f,g)$ be a claw on $[a,b]$, and let $a',b'$ such that $a<a'<b'<b$. Then there are three increasing finite sequences $(x_k)_{|k| \leq M},(y_{1,k})_{|k| \leq M}$ and $(y_{2,k})_{|k| \leq M}$, with
$$
x_{-M} \leq a', \ b' \leq x_{M}
$$
$$
\frac{y_{1,k+2}-y_{2,k}}{x_{k+2}-x_k} \leq 1, f(x_k) \lt y_{1,k} \lt y_{2,k} <g(x_k), |y_{2,k}-y_{1,k}| \lt \epsilon
$$
