If U is uniformly distributed on [0, 1], then the probability that U belongs to some set E is the length (more precisely, measure) of E. In particular, the probability that 0 <= U <= sqrt(x) is sqrt(x) - 0 = sqrt(x).

Why do you think that the graph of a function F(x) = sqrt(x) is supposed to be similar to the graph of its derivative f(x) = 1 / (2sqrt(x))? You can verfy that f(x) = F'(x) and you can plot both functions. They way they look is just a fact.

That's correct.

I am not sure I understand your concern. This is like saying that we have the graph of y = x, but there is never a need to consider the graph of y = x^n because x^n is just a repeated integral of x times a constant.

Provided the two tosses are independent, P(X = 0) = 9/16; P(X = 1) = 6/16 and P(X = 2) = 1/16. Therefore, the cumulative distribution function F(x) of X is 0 for x < 0, F(x) = 9/16 for 0 <= x < 1, F(x) = 15/16 for 1 <= x < 2 and F(x) = 1 for x >= 2.