Finding tension during a tug of war 
Givens:
Mass of Team 1: 612 kg
Mass of Team 2: 657 kg
Force of Team 1: 12150 N
Force of Team 2: 12285 N
Total Acceleration: 0.11 m/s^2
In part b, I've solved for tension using mass of team 1 and force of team 1, the solution was correct which is 12082.68 N, if you use team 2's quantities you also get the same answer. But Newton's second law of motion states that the net external force is equal to mass timesa acceleration. In both of these cases I wasn't using all of my external forces, and I when I tried to use all of them, I'd get the wrong answer. Can anyone explain why?
 A: To understand why you didn't get the right answer we'd really need to see how you included "all the external forces" in the calculation. There is a net force of (12285-12150 = 135)N on the whole system; with a total mass of 1269 kg, this gives an acceleration of $0.11~ \rm{m/s^2}$ as you correctly calculated. 
This is the only calculation that uses "all the external forces". You next calculate the tension by considering one of the teams as a subsystem (at which point the tension becomes an "external force" on that subsystem, as does the force exerted by the ground. The difference between these forces accelerates the team.)
You will have to explain what else you did to "not get the right answer". If you got the tension and acceleration correct, you must have done something like I described. And as far as I can see, that's all there is to it...
UPDATE 
Since you are still having trouble, here is a bit more. First, a diagram:

As you can see, team A has two forces acting on it: $F_1$ to the left, and $T$ to the right; for team B, it's $F_2$ to the right, and $T$ to the left. I will use "to the right" as positive.
Taken as a system, they have force $F_2 - F_1$ acting on it - that's how we calculated the acceleration. For the tension, we can say that the net force on team A must give them the same acceleration $a$ as we got for the whole system; it follows that
$$\begin{align}T - F_1 &= m_1 a \\
&= m_1 \frac{F_2 - F_1}{m_1 + m_2} \\ 
\\
\Rightarrow 
T &= F_1 + \frac{m_1}{m_1+m_2}(F_2-F_1)\end{align}$$
Without putting in the numbers, I can see that the tension will be similar to the force of the ground, which is what you would expect: obviously $\frac{m_1}{m_1+m_2}<1$, and $(f_2-F_1)<<F_1$.
