Even and odd signals
I am not sure that you would call this a practical application as much as a practical implication, but knowing (or forcing) a signal to be odd or even enables you to process it in a certain way.

For example, if you know (or force) a signal to be even, then the imaginary part of its Discrete Fourier Transform (DFT) goes to zero and you can represent this signal just as a sum of $\cos(\cdot)$ functions. This relates the DFT to the Discrete Cosine Transform (DCT).

Specifically:

$$X[k] = \sum_{n=0}^{N-1} x[n] \cdot e^{-i \frac{2 \pi k n}{N}}$$

This is a compact way of saying:

$$X[k] = \sum_{n=0}^{N-1} x[n] \cdot \left ( cos\left ( \frac{2 \pi k n}{N} \right ) - i \sin\left ( \frac{2 \pi k n}{N} \right ) \right ) $$

This makes it clear that $X \in \mathbb{C}$ (omplex) and is composed of a real part (where $x[n]$ is multiplied by the $\cos(\cdot)$) and an imaginary part (where $x[n]$ is multiplied by the $\sin(\cdot)$).

But, $\cos(\cdot), \sin(\cdot)$ are even and odd functions respectively and the way their symmetry works when combined with integration means that if a signal is an even periodic function, then it can be represented by a series of cosines.

OK, but I only have a sound recording that could be anything...how do I know (or force) it to be an even function? (or an odd function).

One of the key assumptions of the Discrete Fourier Transform is that $x[n]$ is a periodic function. That is, even if you pass some random recording of a person counting "one two three four", all that the DFT sees at its input is "one two three four one two three four one two three four one two three four...". In other words, once you reach the length of $x[n]$ (which is $N$), you wrap back to its start.

So, how do we make this an even signal? 

Notice that in even signals $f(x) = f(-x)$. We could therefore generate a new signal that is double the length of $x[n]$ and once we reach the end of $x[n]$ we repeat it by playing it backwards. In this way, $x[N+n] = x[N-n]$ or "one two three fourouf eerht owt eno one two three fourouf eerht owt eno one two three fourouf eerht owt eno..." and so on.

In terms of a "demo", the DFT of [1,2,3,4] has an imaginary component while the DFT of [1,2,3,4,3,2] doesn't.

I do not think that you will find a direct application as such, it will probably be something that "exploits" the implication that a function is even or odd.

Hope this helps.