The object weight here is determined by its displacement I believe which is calculated by its size and the amount of buoyant force required. I entertain someone else helping us out here.
You've got it right. Any object that is floating will be displacing its weight. Take any object that weighs 100 pounds, and as long as it's less dense than water it will be displacing 100 pounds of water once it's immersed and floating. If you know the weight you can calculate displacement. If you can measure displacement you can calculate weight.
For #2 can you explain your math please
It's just junior high algebra. We'll ignore any other factors and only deal with the BCD. Pressure and volume are inversely proportional - as pressure goes up, volume goes down, or vice versa. Since it's a direct relationship the change in one is proportional to the change in the other. Start with an easy example of 1 ATA (the surface, at sea level) and 1 CF of air.
The 1 ATA is P1, and the 1 CF is V1, so P1 * V1 = 1. If you double the pressure the volume will be squeezed twice as hard, so it will be compressed to half of its original volume. That makes P2 2 ATA and V2 .5 CF. 2 * .5 is also 1.
Similarly, if you compress it to 3 ATA the volume will be 1/3 of what it was, so 0.33333 CF. 3 * .33333 = 1
Now let's revisit junior high and solve for V2: P1*V1 = P2*V2 = 1 --> (2*V2) = 1 --> divide both sides by 2 to isolate V2:
(2*V2)/2 = V2 = 1/2 = 0.5.
So, for your original scenario P1 is
2 ATA (I'm assuming, since you say buoyancy increases at 10') and V1 is .33333 CF. P2 is 10' of sea water, but we need to use the same units throughout. Each 33' equals 1 ATM, so 10' of sea water is 10/33 of 1 ATM, or .303 ATM (that's rounded a bit). You also have to add the 1 ATM from sea level pressure, so P2 = 1.303 ATA.
That gives us P1*V1 = 2 * .33333 = .66666 . P2*V2 is also .66666, so 1.303 * V2 = .66666 and once again we solve for V2:
(1.303 * V2)/1.303 = V2 = .6666/1.303 = .512 CF when you've ascended to 10'.
The increase in buoyancy is .512 CF - .333CF, or 0.179 CF. 0.179 CF * 64 lbs/CF = 11.46 pounds (roughly) of additional buoyancy.
Here's an alternate method that is easier in some ways, but may be more challenging if you don't really understand the inverse relationship and working with fractions. If you're still not completely clear on the stuff above and this doesn't make sense off the bat maybe just continue with the one that seems better.
The 1 ATM of pressure at the surface is equivalent to the pressure from 33' of sea water, so we can use [feet of sea water] as our units for describing sea level pressure. That means the 1 ATM of pressure at the surface is 33 feet of sea water. Adding that to the actual depth of 33' means the pressure at that depth is 66 feet of sea water (33' from air and 33' of actual water). Make sense?
Ascending to 10' reduces the pressure to 43 feet of sea water (33' from the air + 10' from the actual water). That means the pressure compressing the air in the BC has decreased from 66 to 43 feet of sea water. Fractions are just ratios, so P2 at 10 is 43/66 of P1 at 33'. Since pressure and volume are inversely proportional, we can invert that fractional pressure to calculate V2 as 66/43 of V1, or 1.54 times as much. 1.54 * .3333 CF = .512 CF.
