Proving property of Modified Rabin Signature
In short words: when you compute things modulo $n = pq$, you are really computing things simultaneously modulo $p$ and modulo $q$. That's the gist of the Chinese Remainder Theorem. So to prove that $a = b \pmod n$, you just have to prove that $a = b \pmod p$ and $a = b \pmod q$.

Modulo $p$, for any $x$ that is not a multiple of $p$, $x^{p-1} = 1 \pmod p$ (this is Fermat's little theorem). Therefore, $x^{k(p-1)} = 1^k = 1 \pmod p$ for any integer $k$. Now, $p$ and $q$ are big primes, so they are odd. It follows that $q-1$ is even, therefore $(q-1)/2$ is an integer. So $x^{(p-1)(q-1)/2} = 1 \pmod p$. The same holds modulo $q$ for the same reason. By the CRT, you get the result modulo $n$.

Note that while Rabin's cryptosystem is defined with $p = 3 \pmod 4$ and $q = 3 \pmod 4$, these properties are not necessary to demonstrate that $x^{(p-1)(q-1)/2} = 1 \pmod n$. It also works when $p = 1 \pmod 4$ or $q = 1 \pmod 4$. The only requirement is that $x$ is relatively prime to both $p$ and $q$, which is exactly what $\mathrm{gcd}(x,n) = 1$ means.