Tension of 2 massless springs joined together 
A cable is formed by joining two massless springs with different spring constants. Spring A has constant k1. Spring B has constant k2. If mass M1 hangs from the cable, is the tension greater on spring A or is it the same on both springs?
 A: Alrighty, what you want to do is consider the free body diagrams about the small point that connects the two springs together. This means, at that point, you replace the two springs with the equivalent forces, i.e. the tensions for each of the springs.
So, right now, your free body diagram should look like a small point with an upward tension (due to spring A) and a downwards tensions (due to spring B).
Now, take Newton's 2nd Law ($F=ma$) for that free body diagram. Because the point you have taken is small, we can say the mass is zero. This means the resultant force is zero.
Therefore, the upwards tension equals the downwards tension. The tensions in both strings are equal.
In fact, if you could have a system that consists of a mass hanging on a long single chain of different springs, the tension will be the same in every spring, provided the springs don't break, and that there is no other external forces being applied (like someone grabbing the centre of the chain).
Interestingly, this is analogous to an electric circuit with a lot of resistors connected in series, where the current is the same in each resistor. Furthermore, if we consider a system where a mass is connected to a ceiling by two springs, where each spring directly connected to the mass and the ceiling, then the weight of the spring is "shared" between the tensions of each spring, which makes sense. This is analogous to a parallel resistor circuit, where the current is shared between each of the parallel branches.
This analogy works to the extent that you could replace the spring system with an electrical circuit containing resistors. This would involve relating:
Tension, $T$ $\iff$ Current, $I$
Extension, $x$ $\iff$ Potential Difference, $V$
Hooke's Law, $T = kx \iff$ Ohm's Law, $I = \frac{1}{R} V$
A: Hard to answer this without just suggesting an answer. 
Consider the forces on the mass. There is gravity that is downwards, but I guess the mass is not moving so think about the forces on the mass - you could also think about the forces the point where the two springs joint together. Hope this is helpful.
A: what you need to do is replace the springs with equal and opposite forces on both ends and do a Free Body Diagram. The answer comes straight out of this.
