Find the orthogonal complement at $L^2[-1,1]$ of the subspace generated by:
a) polynomials with constant term equal $0$;
b) even polynomials.
I think that the answer to the second part is the odd functions (almost everywhere) set in $L^2[-1,1]$, because that's the unique case such that the integral is $0$. Am I correct?
The first one, I have no idea.
For any subspace $V$ we have $V^\perp = \overline{V}^\perp$. Passing to the closure may simplify the problem.
In (b) the closure of even polynomials is the space of all even $L^2$ functions. Indeed, since polynomials are dense in $L^2$, given an even function $f$ we can find a polynomial $p$ such that $\|f-p\|_{L^2}$ is small. Then $q(x)=(p(x)+p(-x))/2$ is an even polynomial approximating $f$.
Concerning (a): the uniform closure of polynomials vanishing at $0$ is the space of continuous functions vanishing at $0$ (use Weierstrass to approximate such functions with polynomials vanishing at $0$). But every continuous function can be changed in a small neighborhood of $0$ at little expense of $L^2$ norm. Therefore, the closure contains all continuous functions, and consequently all of $L^2$.