I think this problem is way too easy for the national round. I saw problems similar to a) with the same difficulty proposed for local rounds. Let $F$ be the antiderivative of $f$ with $F(0)=0.$ Then the function $F(2x)-2F(x)$ has zero derivative so it is a constant function. Since its value for $x=0$ is $0,$ it follows that $F(2x)-2F(x)=0,\forall x \in \mathbb{R}.$ Now take the function $g:(0,\infty) \to \mathbb{R},$ $g(x)= \frac{F(x)}{x}$ and notice that $g(2x)=g(x),\forall x \in (0,\infty)$ and so $$g(x)=g \left( \frac{x}{2} \right)=g \left( \frac{x}{2^2}\right)= \ldots = g\left(\frac{x}{2^n}\right),\forall x \in (0,\infty), n \in \mathbb{Z}^+.$$Passing to the limit in the last relation as $n \to \infty$ and taking into account that $\lim_{x \searrow 0} g(x)=f(0)$ (this follows from the definition of $F'(0)$, which is just $f(0)$), we obtain that $g(x)=f(0),$ so $F(x)=f(0)x, \forall x>0.$ We similarly get $F(x)=f(0)x, \forall x<0,$ so we can say that $F(x)=f(0)x, \forall x \in \mathbb{R}.$ (yes, it is also true for $x=0$) By differentiation, we get $f(x)=f(0), \forall x \in \mathbb{R}$, which is indeed a solution. The answer for a) is therefore any constant function. As for b) just put $f(1)=1$, which implies $f(2^k)=1,\forall k \in \mathbb{Z}$ and put $f(x)=0$ for the rest of the inputs. This function is zero almost everywhere, so we are done.