Let $X=\{v\in L^2([-1,1]):v \text{ is constant a.e.}\}$ be a subspace of $L^2([-1,1])$, characterize $X^{\perp}$.
I don't really know what characterize means, I know that: $X^{\perp}=\{f\in L^2([-1,1]):\langle f,v \rangle=0,\ \forall v\in X\}$.
$$\langle f,v \rangle = \int^1_{-1}fv\ dx = v\int^1_{-1}f,\quad \text{since $v$ is constant.}$$
$$v\int^1_{-1}f=0\quad \text{for every $v\in X$ iff}\quad \int^1_{-1}f=0.$$
So $X^{\perp}=\{f\in L^2([-1,1]):\int^1_{-1}f=0\}$.
Is this enough to characterize the set $X^{\perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?
While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.
Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).