I'm not aware of any general methods, and personally I'm not interested in tricks which are only useful in exams. However, I can explain how to find an expression for your particular example. The way to get better with this kind of exercise is to solve a few of them.
A word which contains exactly one $aa$ has the form $xaay$, where $x,y$ contain no $aa$, $x$ doesn't end with $a$, and $y$ doesn't start with $a$. This suggests first determining how to write a regular expression for the set of words avoiding $aa$.
If a word avoids $aa$, then any two occurrences of $a$ are separated by $b^+$. We can describe all such words by the following infinite regular expression: $$b^* + b^*ab^* + b^*ab^+ab^* + b^*ab^+ab^+ab^* + \cdots.$$
This can be represented by the finite regular expression $b^* + b^*(ab^+)^*ab^*$.
For the sake of our original problem, we need to impose the further constraint that the word not end with $a$. We obtain the following infinite regular expression for the set of all words avoiding $aa$ and not ending with $a$:
b^* + b^*ab^+ + b^*ab^+ab^+ + \cdots = b^*(ab^+)^*.
Similarly, a regular expression for the set of all words avoiding $aa$ and not starting with $a$ is $(b^+a)^*b^*$.
Putting everything together, a regular expression for the set of all words containing exactly one copy of $aa$ is