All strings that contain no run of a's of length greater than two for $\Sigma = \{a,b,c\}$ The solution to this problem is $$(b + c)^*+(b + c)^*((a + aa)(b + c)^+)^*(a + aa)(b + c)^*.
$$
Isn't the $+$ sign a union between sets?, I am asking because I am viewing the line $(b + c)^*+(b + c)^*$
as $A\cup A$ which is $A$ so I do not see a reason to repeat the same set.
The $+$ sign stands for union, but you got the order of operations wrong. Your regular expression is the sum of $(b+c)^*$ and
$$ (b+c)^*((a+aa)(b+c)^+)^*(a+aa)(b+c)^*. $$