Bromination of cyclohexene
To answer this question you have to look at the reaction mechanism. Since the reaction is conducted in the dark, you don't have any radicals to worry about. Under these condition bromine reacts with an electron-rich alkene in a electrophilic addition reaction in the course of which a cyclic bromonium ion is formed. This bromonium ion is highly reactive and in the absence of other nucleophiles reacts with the leftover $\ce{Br-}$ ion in an $\text{S}_{\text{N}}2$ reaction which leads to the opening of the 3-ring and formation of the trans product.



Update:

As @BenNorris pointed out, the ring structure in the question are drawn wrongly. So, I will add the correct version: The dibrominated product will initially be formed in a diaxial conformation (this is always the case due to the ring geometry and the mechanism of the $\text{S}_{\text{N}}2$ reaction). This is usually a very unfavourable comformation so the ring will subsequently flip (if possible) to the diequatorial conformation.