If the answer isn't 5 or 12, you surely enough broke me  

'between >=36 & <50 small dogs' would actually be the correct answer in this case (42.5 is the point between 36 and 49), as the number of large dogs is never stated nor can be worked out; the way the question is worded, there could be 0 large dogs for all anyone knows. 

How I would have solved this as a kid:

49 is the total number. Dividing that number into two groups where one group equals 36, that's all ready a large majority of the total; leaving only 13. Therefore, the "mistake" in this problem is saying "there are 36 more small dogs than large dogs", Those underlined words are unnecessary information, completely misleading; so just drop those words. The problem should have just said "There are 36 small dogs signed up to compete." Therefore, the answer is all ready in the question. How many small dogs are signed up to compete? 36. Basically, a trick question. They want you to say 13, because 13 + 36 = 49. However, 13 is the number of large dogs, and that's not what the question is asking for. 

Blame my schools for making me think like this. We had many teachers that taught multiple subjects, so sometimes things would get mixed up. We would be looking for grammatical puzzles and disregarding unnecessary info before solving math problems like this. And yes, they would also throw in trick questions to catch us off guard.

S = number of small dogs
L = number of large dogs

S + L = 49
S - L = 36

Adding both equations, we get
2S = 85

Divide both sides by 2
S = 42.5
L = 49 - S = 6.5

So there are 42.5 small dogs and 6.5 large dogs. It doesn't make much sense, so maybe we should add another category, "medium dogs". For example, if there is one medium dog, then there are 42 small dogs and 6 large dogs.

(Solved without looking at the video or other posts here.)