1. Let the angular acceleration of the system be β and the angular velocity be ω.

By conservation of energy: m1glcos θ/2+m2 glcos θ = j ω 2/2,

J is the moment of inertia of the system: j = m 1l 2/3+m2l 2.

Then ω = √ (3m1gcosθ+6m2gcosθ)/(2m1l+6m2l).

By: Jβ=m 1glsinθ/2+m2glsinθ, (the first derivative of the moment of momentum of the system with respect to time is equal to the sum of the external moments suffered by the system).

The solution is: β = (3m1gsinθ+6m2gsinθ)/(2m1l+6m2l).

2. When a person rises at a constant speed relative to the rope, the acceleration of a person relative to the weight is zero, that is, the acceleration of the person and the weight is equal.

Let: the acceleration of the weight be: a.

Then it is: ja/r+(m1+m2) ra = (m1-m2) gr (the first derivative of the moment of momentum of the system with respect to time is equal to the sum of the external moments of the system).

m 1=2m2=m,j=(m/4)r^2/2=mr^2/8

Then: MRa/8+3MRa/2=MgR/2, which is simplified as: Ra+ 12Ra=4gR.

Solution: a=4g/ 13.