Formation of hydrogen bromide: Does the concept of "order" apply to this reaction?
We don't assign orders. An order either simply is there, or it isn't. In particular, this expression doesn't look like $\ce{\underbrace{(something)}_{\substack{\text{may include }\ce{[H2]},\\ \text{but not }\ce{[Br2]}}}\cdot[Br2]^a}$ for any $a$, hence there is no order.

Alternatively, if we look at the reaction at its initial moments when $\ce{[HBr]}\approx0$, the denominator simplifies to just $1$, and we may say that the reaction has order $1\over2$ with respect to bromine.

Having or not having an order doesn't have any particularly far-reaching physical consequences. Moreover, it is not even an intrinsic property of a reaction; it may depend on the approximation used, as you saw above.

The scary rate equation comes from the so-called steady state approximation and the detailed reaction mechanism with intermediates and stuff, which is about two times longer than my answer, so I'd better refrain from including it. Well, here is the first line, just to give you the taste of the whole thing:
$$\ce{Br2}\underset{k_{-1}}{\stackrel{k_1}{\longleftrightarrows}}\ce{2Br^.}$$

To tell the truth, most real reactions are like that. The well-known $r=k\ce{[A]^a[B]^b}$ expression with stoichiometric coefficients for exponents is but a great oversimplification. It applies to reactions with no intermediates, which are quite rare.