I lifted this from Ziemer's Modern Real Analysis book, around page 204. The important point is that $f$ and $F^*$ have the same distribution function, namely $F$; this is your problem 4.
Here's what Ziemer says (his notation; not translated into your notation).
Let $\mu$ be a nonnegative Radon measure defined on $\mathbb{R}^n$ and suppose $f$ is a $\mu$-measurable function defined on $\mathbb{R}^n$. Its distribution function, $A_f (·)$, is defined by $A_f (t) := μ(\{x : |f(x)| > t \})$.
The non-increasing rearrangement of $f$, denoted by $f^*$, is defined as $(*) \qquad f^*(t) = \inf \{y : A_f(y) ≤ t\}.$
For example, if $\mu$ is taken as Lebesgue measure, then $f^*$ can be identified with that radial function $F$ defined on $\mathbb{R}^n$ having the property that, for all $t > 0$, $\{F > t\}$ is a ball centered at the origin whose Lebesgue measure is equal to $μ(\{x : |f(x)| > t\})$. Note that both $f^*$ and $A_f$ are non-increasing and right continuous. Since $A_f$ is right continuous, it follows that the infimum in $(*)$ is attained. Therefore if $f^*(t) = y$, then $A_f (y') > t$ where $y' < y$. Furthermore, $f^*(t) > y$ if and only if $t < A_f (y)$. Thus it follows that $\{t : f^*(t) > y\}$ is equal to the interval $(0,A_f (y))$. Hence $A_f (y) = λ(\{f^* > y\})$, which implies that $f$ and $f^*$ have the same distribution function. $\textit{Note: } \lambda$ is Ziemer's notation for Lebesgue measure.
$\ldots$ Notice also that right continuity implies $A_f(f^*(t)) ≤ t$ for all $t > 0$. $<\text{end of extract from Ziemer}>$