Yo-Yo Physics Question? Everyone I asked has been wrong? Please help me?

Did it have a picture? I'm going to guess that "Yo-yo-like" means it can be modeled as a pair of flat disks connected by a stem, rather than as a sphere. No information is given about how wide the stem is, so I'm going to assume it contributes only a very small amount to the mass/moment of inertia and ignore it.

(You were perhaps given these details but you kept them from us, so we're left to guessing).

Under those assumptions, the two halves of the Yo-Yo can be treated together as a cylinder and the moment of inertia is I = (1/2)MR^2. You were given both M and R. R is the radius of the Yo-Yo, 7 m.

B. There may be a shorter way to analyze this, but in problems like this (moving and rotating) as in any other problems, I often fall back on analyzing energy. As it falls, its energy goes into both linear and rotational kinetic energy. How the energy is divided depends on the moment of inertia.
If rotating at angular velocity w, the linear velocity is v = w*r... aha, here's a place you can mess up. Because the cord is wrapped around the 3m step, the radius to use in that relationship is 3 m, not 7m. v = w * 3. That is the speed at which the cord unrolls and therefore the speed at which the whole Yo-Yo is descending.
OK so for any given amount of energy kinetic energy E we have E = (1/2)mv^2 + (1/2)Iw^2
= (1/2)mv^2 + (1/2) [ (1/2)mR^2 ] (v/r)^2
Note big R means the 7 m, which goes into the moment of inertia, and little r refers to the 3 m, which relates v and w.

= (1/2)mv^2 + (1/4)m(R^2/r^2) v^2
= [ (1/2) + (1/4)R^2/r^2] mv^2 = [(1/2) + (1/4)(7^2/3^3) ] mv^2

This is enough to answer part D. The energy E is mg∆h. Set that equal to the above expression and solve for v.

Now how do I use it to answer B? Have to think about that a bit and I'm out of time. Later.

Revisit your other post of this question. I hate to have you "burn" your last shot, but I have a solution posted there. Best of luck.