let $U$ be a random variable uniformly distributed between $0$ and $1$.
A. For any strictly increasing function $f:\Bbb R \rightarrow [0,1]$, find the CDF of $X =f^{-1}(U)$
B. Given any random variable $X$ with PDF $p(x)$, show that $X$ can be generated by $X =f^{-1}(U)$,where
$$f(x)=\int^x_{-\infty}p(v) \, dv$$
What does $X =f^{-1}(U)$ mean? inverse? or just like $5^{-1}=\frac{1}{5}$?
The number $\dfrac 1 5$ is the multiplicative inverse of $5$.
The function $f^{-1}$ is the compositional inverse of $f$.
$5^8$ means $5\times5\times5\times5\times5\times5\times5\times5.$ The operation of multiplication of numbers is iterated through $8$ instances of the number $5$.
$f^8(x)$ means $f(f(f(f(f(f(f(f(x)))))))).$ The operation of composition of numbers is iterated through $8$ instances of the function $f$.
Multiplying by $5^{-1}$ means multiplying $-1$ times by $5$, which amounts to dividing by $5.$
Finding the value of the function $f^{-1}$ at a certain input means applying $-1$ times, the function $f$, i.e. finding the image of that input under the inverse function. (For example, if $f(31)=97$ then $f^{-1}(97) = 31.$)
So, yes, it means inverse, but it's a compositional inverse, not a multiplicative inverse.