A5/2: Ciphertext Only Attack
The idea in [1] (disregarding reception errors) is to first remove the impact of the message. This is done by computing the syndrome

$\qquad\qquad\mathbf{H} \mathbf{v} = \mathbf{H} (\mathbf{m} \oplus \mathbf{k}) = \mathbf{H}\mathbf{k}$.

In the case of a ${1}/{5}$ repetition code, this could correspond to adding even number of (bold) codeword symbols together (since it is a repetition code they should be equal, right?), forming linear equations containing only keystream bits. Then, by exhaustively trying all $2^{16}$ starting states, it is possible to mount an attack (for each starting state) consisting of Guassian elimination, solving for the session key.

I found a reference implemenation at [2], which you might want to check out.

[1] http://www.cs.technion.ac.il/users/wwwb/cgi-bin/tr-get.cgi/2006/CS/CS-2006-07.pdf

[2] http://www.npag.fr/project-a52hacktool