As in the title, say $\lambda$ is some irrep of the symmetric group $S_n$, and $Br_n$ the braid group on $n$ strands,
What is $H^*(Br_n, \lambda)$?
As in the title, say $\lambda$ is some irrep of the symmetric group $S_n$, and $Br_n$ the braid group on $n$ strands,



I claim that the dimension of $H^*(Br_n;\lambda)$ is twice the dimension of the subspace of $\lambda$ fixed by the subgroup $S_2\leq S_n$. Indeed, there is a spectral sequence $$ H^p(S_n;H^q(PBr_n;\lambda)) \Longrightarrow H^{p+q}(Br_n;\lambda). $$ As a $PBr_n$module, $\lambda$ is just a sum of finitely many copies of $\mathbb{Q}$ with trivial action. This means that $H^q(PBr_n;\lambda)=H^q(PBr_n)\otimes\lambda$, which is a $\mathbb{Q}[S_n]$modules and thus injective over $\mathbb{Z}[S_n]$. This means that the spectral sequence collapses to an isomorphism $$ H^*(Br_n;\lambda) = (H^*(PBr_n)\otimes\lambda)^{S_n}, $$ so you just want to know the isotypical pieces of $H^*(PBr_n)$. The classifying space of $PBr_n$ is the configuration space of distinct $n$tuples in $\mathbb{C}$, and there is an elegant argument based on this and the Lefschetz fixed point theorem to determine the character of the $S_n$action on $H^*(PBr_n)$. The answer is that if we take two copies of the trivial representation of $S_2$ and induce up to $S_n$, we get $H^*(PBr_n)$. The claim now follows by Frobenius reciprocity. This tells you the total dimension of $H^*(Br_n;\lambda)$, but not the degrees of the generators. For that, you would need to look in more detail at the structure of $H^*(PBr_n)$. There are generators $\alpha_{ij}\in H^1$ for $1\leq i,j\leq n$ with $i\neq j$, subject only to the relations $\alpha_{ij}=\alpha_{ji}$ and $\alpha_{ij}^2=0$ and $$ \alpha_{ij}\alpha_{jk} + \alpha_{jk}\alpha_{ki} + \alpha_{ki}\alpha_{ij} = 0. $$ 


I'll assume that $\lambda$ denotes an irreducible $\mathbb{Q}S_n$module. I don't know how to compute the dimension of the vector space $H^i(B_n; \lambda)$. But Corollary 4.4 of my paper Representation theory and homological stability with Benson Farb states that for fixed $\lambda$, the twisted cohomology group $$H^i(B_n; \lambda[n])$$ becomes independent of $n$ for $n\geq 4i$. Here $\lambda[n]$ denotes the partition obtained from $\lambda$ by adding boxes to the first row so that it becomes a partition of $n$. This means that even for large $n$, your question can be reduced to a finite computation, as long as the first row of $\lambda$ is much larger than the rest of it. 

