Why are superalgebras so important? I know that a superalgebra is a $\mathbb Z/2\mathbb Z$-graded algebra and that it behaves nicely. I know very little physics though, so even though I know that the super- prefix is related to supersymmetry, I don't know what that means; is there a compelling mathematical reason to consider superalgebras? 
 A: Sure, supermodules over a superalgebra form a nice and nontrivial example of a tensor category. In fact, it is a theorem that Deligne that all tensor categories over $\mathbb{C}$ with suitable growth conditions can be obtained as categories of super-representations.
Another reason is given by Qiaochu, that supercommutativity arises naturally in a lot of places.
For instance, the exterior algebra is supercommutative; thus the wedge product on differential forms acts the same way.  Interestingly, the exterior algebra of a direct sum corresponds to taking the super tensor product.
Another example: With singular cohomology on a topological space, there is a way to define a cup product, which satisfy the supercommutative law $ab = (-1)^{\deg b \deg a} ba$.
(Note that given a $\mathbb{Z}$-graded super-commutative algebra, you can get a $\mathbb{Z}/2$-graded superalgebra as you indicate in your question by taking the sum of the odd parts and the sum of the even parts.)
A: I can summarize one really basic reason, which is actually the reason I originally got interested in the definition.  Take a finite-dimensional vector space $V$ of dimension $n$, and let $\left( {n \choose k} \right) = {n+k-1 \choose k}$ denote the number of multisets of size $k$ on a set of size $n$.  (Multisets are like subsets except that more than one copy of a given element is possible.)  Then the symmetric powers of $V$ have dimensions
$\displaystyle \left( {n \choose 1} \right), \left( {n \choose 2} \right), ... $
whereas the exterior powers of $V$ have dimensions
$\displaystyle {n \choose 1}, {n \choose 2}, ....$
Now here is a funny identity: it is not hard to see that ${n \choose k} = (-1)^k \left( {-n \choose k} \right)$.  One way we might interpret this identity is that the $k^{th}$ exterior power of a vector space of dimension $n$ is like the $k^{th}$ symmetric power of a vector space of dimension $-n$, whatever that means.  So what could that possibly mean?
The answer (and I'll let you work this out for yourself, because it's fun) is to work in the category of supervector spaces!  A supervector space $V$ is a direct sum $V_0 \oplus V_1$, and while one notion of dimension is to take $\dim V_0 + \dim V_1$, another (which can be motivated by thinking of $\mathbb{Z}/2\mathbb{Z}$-graded vector spaces as the category of representations of $\mathbb{Z}/2\mathbb{Z}$) is to take $\dim V_0 - \dim V_1$.  So while a purely even vector space has positive dimension, a purely odd vector space has negative dimension.
Then: graded-commutativity implies that the symmetric power of a purely odd vector space is the exterior power of the corresponding purely even vector space.  More generally, the $k^{th}$ symmetric power of a vector space of dimension $n$ (for all integers $n$) has dimension $\left( {n \choose k} \right)$.
(And, of course, the symmetric algebra of a supervector space is naturally a graded-commutative superalgebra.)
(The physics connection is that symmetric powers = bosons, exterior powers = fermions, and there is a duality between the two.)
A: If you are willing to build a formalism of supergeometry then some constructions are easier to state in this language, notably differential forms and the De Rham complex.  The first one is just functions from the odd line ($R^{(0 | 1)}$) to your manifold, and the De Rham differential comes from super-diffeomorphisms acting on that line.
Also, there is a symplectic/orthogonal duality in representation theory that some people (most famously Kontsevich) have advocated as best understood in terms of Lie superalgebras.
Constructions using a formally negative-dimensional object (or calculations that look like they might come from such an object) sometimes can be interpreted in terms of bona fide $Z/2$ graded objects whose superdimension (even dimension minus odd dimension) is the negative dimension in question.
edit: you will get more knowledgeable answers if you post to Mathoverflow, where some of the contributors have written papers on super- or noncommutative geometry.
