Here is a sketch of how to construct a counterexample satisfying 3':
Fix an enumeration $\mathbb{Q}=\{q_n :n \in \mathbb{N}\}$ and let $g:\mathbb{R} \to \mathbb{R}$ be the function defined by $g(x)=\sqrt{x-\pi}$ for $x \geq \pi$ and $g(x)=-\sqrt{\pi-x}$ if $x<\pi$. It doesn´t really matter what $g$ is as long as it satisfies the following properties:
a) $g$ is a continuous increasing bijection and $g(\pi)=0$.
b) For any rational $q$ and any $\epsilon>0$, there are irrationals $l < q$ and $r > q$ such that $r-l < \epsilon$ and the affine function sending $[l,r]$ into $[g(l),g(r)]$ has rational coefficients.
c) $g$ is unbounded from above and from below.
Now one can define inductively a disjoint sequence of intervals $\{ [l_n,r_n]\}$ with irrational endpoints such that for each $n$ the affine function (lets call it $h_n$) sending $[l_n,r_n]$ into $[g(l_n),g(r_n)]$ has rational coefficients. At each step, make sure that $\pi \notin [l_n,r_n]$ and that $q_n \in (l_m,r_m)$ for some $m \leq n$.
The function $f: \mathbb{R} \to \mathbb{R}$ defined as $f(x)=h_n(x)$ if $x \in [l_n,r_n]$ and $f(x)=g(x)$ otherwise is continuous, increasing, unbounded and the function $f \upharpoonright \mathbb{Q}$ satisfies conditions 1,2,3' and 4. However $f(\pi)=0$.