Buoyant force physics? I have been studying for 7 hours straight and still don't understand it

ok here's the question
a cube of wood having and edge dimension of .2 m and a density of 650 kg/m^3 floats on water.
a) what is the distance from the horizontal top surface of the cube to the water level
b) what mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface?

for question a i tried to work it out and got the right answer but I don't think I understand the concept.
first I use the density of wood and multiply that by the volume of wood to get mass of the wood.
(.2m x .2m x .2m) (650) = 5.2 kg

Now I use the formula Fbuoyant(rho x V of displaced liquid x g) = mg and I know that Fbuoyant is the same as weight of displaced liquid. Now, IS THE WEIGHT OF THE DISPLACED LIQUID THE SAME AS THE WEIGHT OF THE OBJECT?

I went ahead and did
Rho x V of displaced liquid x g = m g
all the g's cancel and I am left with
rho x V of displaced liquid(l x w x h) = 5.2 ( i am not sure if I am supposed to plug in this 5.2 because I don't know if the mass of displaced water is the same as mass of object?)
1000 x .2 x .2 x h = 5.2
then solving for h I got = 0.13. Now I don't even know what this h means
but I played around with the numbers and subtract that from the original height of the object of .2m and got .7 m. It is correct but i still don't understand.

part b)
I used Fbuoyant =weight of displaced water
Again I don't know if weight of displaced water is the same as weight of objects but I substituted it anyway.
rho x V displaced liquid x g = m g block + m g lead
g cancels so:
1000 x .008(now i thought volume of displaced liquid is NOT the same as volume of object if the object isn't completely submerged? but i plugged it in anyway). = 5.2 + m lead
so mass of lead = 8 - 5.2 = 2.8 kg
It is right but I still don't get it.
Please help :(

If you have been studying for 7 hours, take a break, get some fresh air and then read the explanation below.