@Lopsy  
Oh, that’s nice.  
But I don’t think they’d require a proof for this. The first one is too simple.

If this is an olympiad, they probably want a proof for #2. Here’s a quickie:
 
Consider a rotation centered at A which rotates everything clockwise 60 degrees. Since ABC’ and AB’C are both equilateral triangles, this rotation takes B to C’, and it takes B’ to C.
 
Therefore, this 60 degree rotation takes line BB’ to line C’C. So, the directions of the lines BB’ and C’C are 60 degrees apart.
 
This means that angle B’OC is 60 degrees. So, angle BOC is 180-60 = 120 degrees.
 
So the answer is always 120 degrees, no matter what the original triangle looks like.

I use to think like this in problems like the second one.  
It looks like we are either missing some data, or the solution is independent of such data.
 
Assuming the problem makes sense, the latter hypotesis holds. So we can build ABC in the most simple way for our calculations.  
In this case, just make B and C coincide: now it’s trivial to get that C’OB’ is 120 degress, with O coinciding with B and C.
 
The first one doesn’t even need to be commented.

@Sparkler
 
And again, I want to edit my own comments. I just got what ‘altitude’ means and why that reuires 120 degrees as an answer.

n = 21  
12n^2 + 12n + 11 = 5555
 
120 degrees.

@Ilsor
 
They’re probably looking for an answer in the form of a trigonometric function.

@Ilsor  
That’s how I got it, actually.

@Background Pony #700F  
You’re right.
 
Since we are given absolutely no information about the original triangle ABC, we can assume that the answer remains the same for any triangle ABC. If the answer would change depending on attributes of ABC, we would have been given more info about ABC, like angles or whatever.
 
So, let’s make ABC an equilateral triangle.
 
If we do that, CC’ and BB’ suddenly become the altitudes of ABC, giving us the answer of 120 degrees.

@Sparkler  
It’s okay. I misread it too. It should be 120. But still, I might very well be wrong.

Ah, you’re right, I misread the question as saying both ABC and AB’C - I didn’t see the mark for C’.

@Sparkler  
I think you got the second wrong.
 
I’m not sure but I think the answer is 60 degrees. I might very well be wrong, though.

@Sparkler  
Regarding the second problem:  
There is no common border in the problem.

Why can’t I pay to have an ‘edit’ button? of course I meant 12(n^2 + N).

@Ridley Wolf  
You mean ∞ ³?

Well… the first question’s easy. It just told you there are only ten possible values that it can be, from 1111 to 9999. Subtract 11 from each, gettin 1100 to 9988. Divide the values by 12, since the total is 12(n^2 +). Only one will divide by 12.
 
The second one’s easier. The problem tells you two identical triangles share a common border. That means that the line that links their points (BOC) is perpendicular to the line that forms their common border. A perpendicular angle is 90 degrees.

The answer, of course, is ∞²/0.