One of my first thoughts is to try to derive $(1)$ from $(2)$ by integrating by parts: $ \int_0^\infty xf(x)\,dx = \int x\,dv = xv-\int v\,dx = \left.xF(x)\right|_0^\infty - \int_0^\infty F(x)\,dx\ldots \text{?} $ This won't work, for reasons you'll see if you think about what that last thing says. So go from $0$ to $a$ and then take $\lim\limits_{a\to\infty}$: $ aF(a) - \int_0^a F(x)\,dx = \int_0^a F(a)-F(x)\,dx = \int_{(0,\infty)} 1_{(0,a]}(x) (F(a)-F(x)) \,dx. $ You can then use the monotone convergence theorem to get $ \int_0^\infty 1-F(x)\,dx. $ as the limit.
(Of course then you'd want to show $(2)$ follows from $\mathbb EX = \int_\Omega X\,dP$.)
Later addendum: Let $Y=X^q$. Then if you've established $(1)$, then you have $ \mathbb EY = \int_0^\infty P(Y>y)\,dy. $ Let $y=x^q$ so that $dy=qx^{q-1}\,dx$. And notice that $Y>y$ iff $X>x$, so that $P(Y>y)=P(X>x)$. Then the integral above becomes $ \int_0^\infty P(X>x)\left(qx^{q-1}\,dx\right). $
Still later addendum: How to prove $(2)$ is something I'd forgotten the details of.
We have a probability measure $P$ on measurable subsets of a space $\Omega$, and a measurable function $X:\Omega\to\mathbb R^+$. "Measurable" will mean that inverse-images of Borel subsets of $\mathbb R$ are measurable subsets of $\Omega$. The "probability distribution" of $X$ is a probability measure $Q$ on Borel subsets of $\mathbb R^+$, defined by $Q(A) = P(\{\omega\in\Omega : X(\omega)\in A\})$. So the problem is to prove that $ \int_\Omega X(\omega)\,P(d\omega) = \int_{\mathbb R^+} x \, Q(dx).\tag{i} $ The integral on the left in $(i)$ is the supremum of integrals of simple function (nonnegative functions with finite images) that are $\le X$. The one on the right is the supremum of integrals of simple functions of $x$ that are $\le x$. It remains to show that there is a bijection between simple functions of $X$ that are $\le X$ and simple functions of $x$ that are $\le x$, and that that bijection preserves the values of the integrals. You therefore get the same set of values of integrals of simple functions in both cases, hence the same sup in both cases.