Does there exist a measurable $A \subseteq [0,1]$ with $\lambda(A \cap [0,a]) = a/2$ $\forall a \in [0,1]$? does there exist a Borel set $A \in [0,1]$ such that, for any $a \in [0,1]$, the Lebesgue measure of the set $A \cap [0,a]$ is $a/2$?

Thanks
If so this would force $\lambda(A \cap [a,b]) = \dfrac{b-a}2$ for every interval $[a,b] \subset [0,1]$.

Whenever $\{I_k\}$ is a cover of $A$ by closed subintervals of $[0,1]$ this would imply $$\lambda (A) = \lambda \left( A \cap \cup I_k \right) \le \sum_k \lambda (A \cap I_k) \le \frac 12 \sum_k \ell(I_k)$$
and by taking the infimum of all such coverings, $\lambda(A) \le \dfrac 12 \lambda(A)$ which forces $\lambda(A) = 0$.

tl;dr NO
Let $f(a) = \int_0^a 1_A$, then the Lebesgue differentiation theorem gives
$f'(a) = 1_A(a)$ for ae. $a$. In particular, we can't have $f'(a) = {1 \over 2}$.