Attempting Advanced Problem 6:
##X## and ##Y## are independent stochastic variables which both have uniform distrubution on ##[0,1]##. This means that for every Lebesgue mesaurable set ##A\subseteq \mathbb R^2##, ##P((X,Y)\in A)=m(A\cap[0,1]^2)##, where ##m## is the Lebesgue measure on ##\mathbb R^2##. Also, for every Lebesgue measurable set ##B\subseteq \mathbb R##, ##P(X^Y\in B)=m(\{(x,y)\in[0,1]^2\,|\,x^y\in B\})##.
The distribution of ##X^Y## is determined by its cumulative distribution function ##F:\mathbb R \to[0,1]##, given by ##F(t)=P(X^Y\le t)=m(\{(x,y)\in[0,1]^2\,|\,x^y\le t\})##, for all ##t\in\mathbb R##. To find the distribution, it should suffice to find ##F##.
The function ##f(x,y)=x^y##, defined on ##[0,1]^2## (for definiteness, we define ##f(0,0)=0^0=1##, but this does not really matter) is increasing in ##x## and decreasing in ##y##. ##f## is continuous on all ##[0,1]^2## except at ##(0,0)##, so it is Lebesgue measurable. We have ##f(x,0)=1## and ##f(x,1)=x##, for all ##x\in[0,1]##, and ##f(0,y)=0## and ##f(1,y)=1## for all ##y\in\,]0,1]##, so the range of ##f## is ##[0,1]##.
Hence, ##F(t)=0## for ##t<0## and ##F(t)=1## for ##t>1##.
For ##t\in[0,1]## and ##(x,y)\in[0,1]\times\,]0,1]##, we have ##x^y\le t\Leftrightarrow x\le t^{\frac1y}(\le 1)##, since ##\frac1y\ge 1##. For each ##\epsilon\in\,]0,1[\,##, we then have ##m(\{(x,y)\in[0,1]\times[\epsilon,1]\,|\,x^y\le t\})=\int_\epsilon^1 t^{\frac1y}dy##. This tends to both ##F(t)## (since ##m(\{(x,0)\,|\,0\le x\le 1\})=0)## and ##\int_0^1t^{\frac1y}## (since this improper integral converges, because the integrand is continuous and bounded on ##]0,1]\,##) as ##\epsilon\to 0##. Hence ##F(t)=\int_0^1 t^{\frac1y} dy##.
I don't know of any closed expression for this integral, and I doubt that such an expression exists or is known.
So my answer is that the desired distribution is given by the cumulative distribution function:
F(t)=\begin{cases} 0 & t<0,\\ \int_0^1 t^{\frac1y}dy,& 0\le t \le 1, \\1,&t>1.\end{cases}
