How to avoid arithmetic mistakes?
There are certain quick methods called sanity checks which will catch most (but not all) arithmetic errors. One common one is to replace each number with the sum of its digits, which is the "casting out nines" method mentioned in Robert Israel's answer. To check a computation, say $567\times 894=506,898$, we replace $567$ with $5+6+7=18$ and $894$ with $8+9+4=21$, and then replace each of these with the sum of their digits to get $9$ and $3$ (in general we keep doing this until we get down to $1$ digit), while on the other side we get $5+0+6+8+9+8=36$ and then $3+6=9$. We then check that $9\times 3=9$ after casting out nines on both sides, and so our answer is probably right (though not necessarily). This method is called "casting out nines" because it ensures that whatever answer you are checking differs from the actual answer by a multiple of nine (hopefully $0\times 9$).
However, this method has a serious drawback: if the answer you are checking is correct except for having the digits switched around (a relatively common error) the method will not catch the error. A remedy for this is to use "casting out elevens" where you take the alternating sums of the digits instead of the sums, such that the last digit is always added rather than subtracted. In our previous example, this becomes $5-6+7=6$, $8-9+4=3$ and $5+0-6+8-9+8=-4$. Here we have to be a little careful: we want to take the equation $6\times 3-(-4)=0$, cast out elevens (take the alternating sum) and verify that the the resulting equation holds, which in this case it does. We move everything to one side so that we don't have to work with numbers of different signs ($6\times 3=-4$ is true $\bmod 11$, which is what matters, but it is not obvious how to cast out elevens to show this). This ensures that whatever answer you are checking differs from the actual answer by a multiple of eleven (hopefully $0\times 11$), hence the name.
Edit: These methods can both be made rigrous with modular arithmetic. The first simple checks that an equation holds $\bmod 9$, and adding the digits comes from the fact that $$\begin{eqnarray} d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\ &\equiv& \sum\limits_{i=1}^n 1^id_i (\bmod 9)\\ &=&\sum\limits_{i=1}^n d_i \end{eqnarray}$$ while the second checks that an equation holds $\bmod 11$, and the alternating sum of the digits comes from the fact that $$\begin{eqnarray} d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\ &\equiv& \sum\limits_{i=1}^n (-1)^id_i (\bmod 11) \end{eqnarray}$$
Credit where credit is due: I believe I read about this years ago in a question to Dr. Math from an elementary school teacher who had been teaching the method and wanted to know how it worked.
In my experience, the best way to avoid computational errors is to avoid computation. Develop general algorithms for whatever quantity that you are looking for and then proceed to "plug and chug" as the last step. Mathematics requires precision, however, and you often cannot avoid having to comb over your work tediously.
I have found the best way to avoid these types of computational mistakes is:
To have extremely neat and clear handwriting.
To effectively use the space on the page to organize the work in a logical manner.
Use a pencil. Never cross things out, but erase them instead.
Always take the time to think things through slowly and carefully.
Keep your desk very neat and well organized.
I believe that if you write things down in a clear way, then you will think in a clear way; and if you write in a sloppy way, you will think in a sloppy way.