Boundary conditions confusion

When finding the particular solution to a second order homogeneous/non-homogeneous differential equations, I keep getting the boundary conditions wrong. I usually get the first one right (regarding displacement), but whenever I try to find the second boundary question relating the velocity to the time at t = 0s, I get it wrong. Here are some examples.
1. https://ibb.co/ZpB7VM3k - I understand the first boundary condition. When t = 0, x = 0. However, the second boundary condition is, when t = 0, dx/dt = U. How can that be possible, when the spring initially lies at rest?
2. https://ibb.co/VhvxWhy - For the second boudary condition, the answer states that when t = 0, dx/dt = 0. However, the question says that "at time t = 0, the lift starts to move vertically upwards with constant speed U".
Am I perhaps thinking in the wrong way?

For 1), its the extension in spring thats the subject of the ode, not the particle, P. displacement. So initially P (A) is at rest, and at time zero, the spring is at its natural length (extension is zero) and B is moving with velocity u, so the extension has initial velocity u as extension is B-A (assuming they refer to positiions) and that gives u-0 so u.
2) is the displacement of the particle, so the opposite of the first case. As the particle is at rest, the initial velocity is zero.

Okay. I think I understand your answer to 2) but for 1), because "initially the spring lies at rest on the horizontal surface" (I don't think I read the question properly), wouldn't that mean that at t = 0, dx/dt is equal to 0?

The extension x(t) is
x(t) = B(t) - A(t) - l
where B is the position of B, similar for A and l is the natural length. If its initially at rest, then B(0)-A(0) = l so x(0)=0. "The end B is then moved .... with constant speed U". So when the motion starts, B has (just) started moving with constant speed U so
dx(0)/dt = dB(0)/dt - dA(0)/dt = U - 0
as A is instantaneously stationary at t=0. Youre saying that B begins move with constant speed U at t=0 and A does not instantaneously respond, due to the dynamics of the spring.

Okay I think that makes sense thank you! The equations you added helped a lot.