for the second question,

 

Clearly for x<=0 there's no soln

For x>=1, also there's no soln (this can be got by grouping the terms)

It suffices to prove that there exists no soln for x belonging to (0,1).

Group the terms as (x^4-x^5) + (x^2-x^3) + (x^6-x) + 3/4 = 0

Now clearly the first 2 terms are >0.

To find the minimum of x^6-x:

Differentiate

implies 6x^5=1 implies x^5 = 1/6

implies x = (1/6)^1/5 implies x^6 - x = (1/6)^6/5 - (1/6)^1/5

>-3/4

 

Hence the result