To solve the first equation, 2lg^2x - 5lgx - 7 = 0, we can use substitution to simplify the equation.

Let y = lgx.

The equation then becomes: 2y^2 - 5y - 7 = 0.

Now we can solve this quadratic equation by factoring or using the quadratic formula:

(2y + 1)(y - 7) = 0
y = -1/2 or y = 7

Now substitute back in lgx:

lgx = -1/2 or lgx = 7

This gives us two possible solutions for x:

x = 10^(-1/2) or x = 10^7
x = 0.316 or x = 10,000,000

Therefore, the solutions to the first equation are x = 0.316 or x = 10,000,000.

For the second equation, log₀,₅(x) + 3logx(0,5) = 4, we can first simplify the equation using logarithmic properties:

log₀,₅(x) + logx(0,5)^3 = 4
log₀,₅(x) + logx(0,125) = 4
log₀,₅(x) + logx(1/8) = 4
log₀,₅(x * 1/8) = 4
log₀,₅(x/8) = 4

Now we can convert this equation to exponential form:

₀,₅^(4) = x/8
16 = x/8
x = 128

Therefore, the solution to the second equation is x = 128.