2 men are selected from 8 men in $^8C_2$ ways.
Since no husband and wife should be in same game,
two women out of remaining 6 are chosen in $^6C_2$ ways.
Now one team can be chosen as $(M_1,W_1)\:or\:(M_1,W_2)$ in 2 ways.
$\therefore$ The required no. of arrangements = $^8C_2\times ^6C_2\times 2$