Realizing Stiefel-Whitney classes via vector bundles Let $X$ be a CW complex. If $E$ is a vector bundle over $X$, then it's well-known that the Stiefel-Whitney classes $w_j(E) \in H^j(X,\mathbb F_2)$ of $E$ are determined from the classes $w_{2^k}(E)$ (for $2^k \leq j$) via the Wu formula, using the cup product and the action of the Steenrod algebra.

Question 1: Does the Wu formula imply any further relations? This is a purely algebraic question which I make more precise in (a) and (b) below.
That is, let $H$ be a nonnegatively-graded $\mathbb F_2$-algebra with an unstable action of the Steenrod algebra satisfying the Cartan formula and $Sq^{|x|}(x) = x^2$ for all homogenenous $x \in H$. Let $W$ be the set of sequences $(w_j \in H^j)_{j \in \mathbb N}$ with $w_0 = 1$ and satisfying the Wu formla.
(a) For any sequence $(v_{2^k} \in H^{2^k})_{k \in \mathbb N}$, does there exist $w \in W$ (necessarily unique) with $w_{2^k} = v_{2^k}$ for all $k \in \mathbb N$?
Presumably (i) the Whitney sum formula and (ii) the universal formula for the Stiefel-Whitney classes of a tensor product of vector bundles are compatible with the Wu formula, so that $W$ is a commutative ring using (i) for addition and (ii) for multiplication.
(b) Is $W$ a polynomial algebra on whichever generators from (a) do exist?

Question 2: What restrictions beyond the Wu formula are there restricting the Stiefel-Whitney classes of a vector bundle $E$ on a CW complex $X$? This is a genuinely topological question.
Over here Mark Grant describes one such restriction, but ideally I'd like a more systematic discussion.

If it simplifies matters to assume that $X$ is finite, or even a compact manifold, then that's fine.
 A: Here is a start at answering Question 1: there are indeed further relations between the $w_{2^k}$, or at least conditions on the $w_{2^k}$.
For example, consider the case where $H$ has multiplication which is null except for what is implied by the multiplication being unital. (For example, $H$ may be the cohomology of a suspension space.)
In this case, the Wu formula reduces to
$$Sq^i(w_j) = \binom{j-1}{i} w_{i+j}$$
So if the $w_{2^k}$'s are given, we are forced to define $w_{2^k + j'} = Sq^{j'} w_{2^k}$ for $0 \leq j' \leq 2^k-1$, which gives us the definition of each $w_j$. So now in the case where $j = 2^k+1$ and $1 \leq i \leq 2^k - 1$, the Wu formula stipulates that $Sq^i Sq^1 w_{2^k} = 0$. This is always the case for $i = 1$, but for all other $i$, the relation $Sq^i Sq^1 = 0$ does not hold in the Steenrod algebra, so I believe there are examples of $H$'s and $w_{2^k} \in H^{2^k}$ where this equation does not hold. So this is an example of some kind of further condition which $w_{2^k}$ may be required by the Wu formula to satisfy.
