Here are the solutions to the given inequalities and trigonometric equations:

Tgx > -1/(√3):
This inequality means that the value of tgx (tangent of x) should be greater than -1/√3. Therefore, x can take any value within the range of (-π/6, π/6) U (5π/6, 7π/6) U (11π/6, 13π/6).

2sinx + √2 > 0:
To solve this inequality, we need to isolate sinx first:
2sinx > -√2
sinx > -√2/2
x can take the values in the range of (-3π/4, -π/4) U (π/4, 3π/4).

√2cosx - 1 = 0:
cosx = 1/√2
x = π/4

сtgx - √3:
I'm assuming there is a missing inequality or equation here.

arccos(-2/2) - arcsin(-1):
arccos(-1) - (-π/2)
π/2 + π/2
π

arctg1/(√3) + arctg√3:
This simplifies to:
π/6 + π/3
π/2

tgx > -1/(√3):
We already solved this inequality in the first point.

I hope this helps! Let me know if you need further clarification.