If only conductive heat loss was significant for a house, then by what percentage would you lower the heat loss if the temperature was reduced inside from 70 f to 60 f when the outside temperature is 20 f?

Let's try and solve this using the heat conductivity equation.Q=−kA(ΔT)/(Δx)

Where Q represents heat loss, k represents thermal conductivity, A represents area of the wall, and x is the thickness of the wall.
We'll assume everything except Q and delta T is for this case only, 1.
Case 1 : 70 F - 20 F = 50 F.Q=−1×1(50)/(1)
That's -50. Negative because energy is being transferred outside.Assuming the same trend..
Case 2 : 60F - 20F = 40F difference.I'm going to skip the calc. for this one, since it's pretty obvious. You're ending up with -40 as the answer here.
Now, let's express the answer of Case 2 where the temp difference is -40, as a percentage of Case 1. 
−40−50×100=80

Therefore, you're lowering the heat loss by 20%. 

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.

A., d., and f i think