The expresssion that you put for the Keq of the A + B <--> C reaction is correct. You're almost right with the Keq for the chlorine/bromine reaction as well, and you've got the right idea with the exponents. However, take a good look at what your products are. The equation you've written is reverse of the way you'd normall see it, i.e., the normal way to write this is:
Br2 + Cl2 <--> 2 BrCl
Why? Well, the initial concentration of BrCl is zero, so it is being formed, thus we consider it to be the 'products' even though the whole process is in equilibrium. If the initial concentration of the bromine and chlorine gasses were zero, then they would be considered the products.
You also ask about the exponent. You did correctly place the coefficient as an exponent on the BrCl term. You can think of this as a simplified reaction; all the products are multiplied together, so you can take this equation:
A + B <--> 2 C
and re-write it as:
A + B <--> C + C
Now you can see where the exponents come from. The Keq would look like this:
[ C ][ C ]
Keq = [ A ][ B ]
Of course, you can simplify that into:
[ C ] 2
Keq = [ A ][ B ]
So, your equation is going to look like this:
[ BrCl ] 2
Keq = [ Br2 ][ Cl2 ]
Remember that you're looking for the total concentration of BrCl, not double, so be careful not to double it, as I see you had 2BrCl written in there. It is true that there are 2 mols of BrCl to one of either Br2 or Cl2, but don't figure out what the concentration of BrCl is and then double it, or you'll wind up with a 4:1 molar ratio instead of the 2:1 that you're looking for. I hope that wasn't too confusing. Now that you've got the equation, you can just substitute in the appropriate algebraic expressions (x, 2x, whatever) and you'll have a nice quadratic equation to solve. If you have any more questions, please drop us another post.