1) We can rewrite the equation as:

6cosx + 5sin^2x - 6 = 0

Since sin^2x = 1 - cos^2x, we can substitute it in:

6cosx + 5(1 - cos^2x) - 6 = 0
6cosx + 5 - 5cos^2x - 6 = 0
-5cos^2x + 6cosx - 1 = 0

Then we can use the quadratic formula to solve for cosx:

cosx = (-6 ± sqrt(6^2 - 4(-5)(-1))) / (2*(-5))
cosx = (-6 ± sqrt(36 - 20)) / -10
cosx = (-6 ± sqrt(16)) / -10
cosx = (-6 ± 4) / -10

So the solutions for cosx are:

cosx = (-6 + 4) / -10 = -2 / -10 = 1/5
cosx = (-6 - 4) / -10 = -10 / -10 = 1

Therefore, x = arccos(1/5) and x = arccos(1).

2) To solve the inequality:

2 log(1/9) 2 - 3x/x > -1

First, simplify the left side using properties of logarithms:

2 log(1/9) 2 = 2log1/9 + 2log2 = log((1/9)^2) + log(2^2) = log(1/81) + log(4) = log(4/81)

Then the inequality can be written as:

log(4/81) - 3x/x > -1

Combine the logarithms:

log(4/81) - 3 > -1

Rewrite in exponential form:

4/81 - 3 > 1

Solve for x:

4/81 > 4
81 > 1

This is always true, so there are no restrictions on x for the inequality to hold. Therefore, all real numbers are solutions.