How could a cord withstand a force greater than its breaking strength?

Breaking strength refers to the maximum tension in the cord. Now, from the sounds of this problem, you've probably been doing force diagrams involving cords. What happens when you attach two cords to a single 100N object (and keep it stationary)? Is the tension in both of those cords 100N? Or is the combined force 100N, so that each just has 50N?

Put another way, most ropes you see will be made of many individual little threads. Each one of them is much weaker than the whole rope. See what I'm getting at?

You have to accelerate the object towards the ground. (Let it fall a bit.) This creates a bit of "slack" in the cord so that it doesn't break. Figuring out how much acceleration it should have is a good exercise.

EDIT: I figure I might as well work it out since this question has so many views. Note that personally I view doubling the rope as cheating. From the free body diagram, we get, in an obvious notation $$W-T=ma$$ We want to find the minimum acceleration needed, so we plug in the max tension, $80\,\text{N}$: $$100\,\text{N}-80\,\text{N}=20\,\text{N}=ma$$ Assuming we are on Earth, the mass of the package is $$m=\frac{W}{g}=\frac{100\,\text{N}}{9.8\,\text{m/s}^2}=10.2\,\text{kg}$$ From above, we get $$a=\frac{20\,\text{N}}{m}=\frac{20\,\text{N}}{10.2\,\text{kg}}=1.96\,\text{m/s}^2$$

As far as I see, if you want the object to lower at constant speed the only way is actually fixing at least both the extremities of the cord to the object (can't describe this well, let's say your "grip" is on the middle of the cord).

Whatever happens somewhere else, constant speed means that the balance of the forces on the object must be 0, so if you tie it in two points you have $F_g + T + T =0$ (being $F_g$ the gravitational pull on the object and $T$ the tension on the cord, with appropriate signs). With this configuration you get $| T | = | F_g | /2 = 50N$, less than the breaking strength.