How could a cord withstand a force greater than its breaking strength? How could a 100 N object be lowered from a roof using a cord with a breaking strength of 80 N without breaking the cord? 
My attempt to answer this question is that we could use a counter weight. But I don't really understand the concept behind counterweights so I hope someone can clear that up for me and if there is a better answer I'll love to know it. 
 A: 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.
A: I would not be surprised if a cord with a given breaking strength of 80N held 100N once. It should be regarded as trash after that though.
You see, they set the breaking strength as guaranteed to hold under the worst-case setup. Select a low-reduction knot (the rating not is the double figure-eight which is common but not the optimal choice) and apply the load smoothly, and you will find the real breaking strength is higher than rated.
It would be unwise to depend on this though.
EDIT: The double figure eight's coefficient is .75 which means the straight line breaking strength is 106N. If we assume 0 margin (as we do in homework problems) this solution requires a coefficient of .943. This means long splice is the only "knot" with a good enough coefficient (the coefficient of the long splice is not known but estimated at 1). While we can make a loop with a long splice, this is really suboptimal.
A: Use a ramp, an incline of 53° will work.
Otherwise You need to double up the cord.
The third option is to just carry the 10 kg object down the stairs.
A: The simplest approach (and what the person asking this question probably was getting at) is to use a pulley like so:

The weight of the object is now shared between the two sides, with each carrying a 50 N load. You end up using twice as much cord. The other advantage is that you now have a "mechanical advantage" and you only need to use a force of 50 N to lower the object. You do need a point where you fix the other end, unless you hold both sides in your hands. In that case the load is evenly balanced between the two halves of the cord by the pulley.
Alternatively you could simply double up the cord - but the tricky thing there is to ensure that they share the load evenly. This is done in practice by twisting the ropes together (yes, that is one reason why ropes since time immemorial are twisted): if one strand carries a larger fraction of the load it tends to straighten out - which makes the other strand "take a longer path" (it becomes more twisty) until it starts carrying more of the load. In this way, twisting ensures sharing of the load. Twisting of the strands also makes a rope more flexible (since strands spend "equal amounts of time" on the inside of the bend and the outside - I put that expression in quotes since it is only approximately true but you get the idea).
A: 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?
A: 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
$$a=\frac{20\,\text{N}}{m}=\frac{20\,\text{N}}{10.2\,\text{kg}}=1.96\,\text{m/s}^2$$
