1) To solve the equation log2(2x+1) = 3, we can rewrite it in exponential form: 2^(3) = 2x+1. Since 2^3 = 8, we get 8 = 2x+1. Now, solve for x: 2x = 8 - 1 = 7, x = 7/2.

2) To solve the equation cos(2x) = cos(x) - 1, we can use the double angle formula for cosine: cos(2x) = 2cos^2(x) - 1. Substituting this into the equation gives us 2cos^2(x) - 1 = cos(x) - 1. Simplifying this equation leads to 2cos^2(x) = cos(x). This results in cos(x)(2cos(x) - 1) = 0, so either cos(x) = 0 or 2cos(x) - 1 = 0. Solving for x, we get x = π/2 + πn, x = 2πn, and x = 2π+π/3 + 2πn.

3) To solve the equation log0.5(x^2 - 5x - 4) = -1, first rewrite it as an exponential equation: 0.5^(-1) = x^2 - 5x - 4. Since 0.5^(-1) = 2, this gives us x^2 - 5x - 4 = 2. Rearranging this equation leads to x^2 - 5x - 6 = 0, which can be factored as (x-6)(x+1) = 0. Thus, x = 6 or x = -1.

4) The equation log2(2x - 1) = log2(3x - 2) = 0 implies that 2x - 1 = 3x - 2 = 2^0 = 1. Solving this system of equations leads to x = -1.