49. Since $x, y, z \geqslant 0$, and $x+y+z=1$, therefore, $0 \leqslant xy, yz, zx < 1$, thus, $\frac{1}{1-xy} + \frac{1}{1-yz} + \frac{1}{1-zx} \geqslant 3$. Below, we prove that $\frac{1}{1-xy} + \frac{1}{1-yz} + \frac{1}{1-zx} \leqslant \frac{27}{8}$. Noting that $x+y+z=1$, the inequality is equivalent to:
$$\begin{array}{l} \frac{3-2(xy+yz+zx)+xyz}{1-(xy+yz+zx)+xyz-x^2y^2z^2} \leqslant \frac{27}{8} \Leftrightarrow \\ 11(xy+yz+zx) + 27x^2y^2z^2 \leqslant 3 + 19xyz \end{array}$$

By Schur's inequality, we have
$$(x+y+z)^3 - 4(x+y+z)(yz+zx+xy) + 9xyz \geqslant 0$$

which means $1 - 4(yz+zx+xy) + 9xyz \geqslant 0$, hence
$$xy + yz + zx \leqslant \frac{1 + 9xyz}{4}$$

To prove (1), it suffices to prove
$$11\left(\frac{1 + 9xyz}{4}\right) + 27x^2y^2z^2 \leqslant 3 + 19xyz \Leftrightarrow 108x^2y^2z^2 + 23xyz \leqslant 1$$

By the AM-GM inequality, we have
$$xyz \leqslant \left(\frac{x+y+z}{3}\right)^3 = \frac{1}{27}$$

Therefore,
$$108x^2y^2z^2 + 23xyz \leqslant 1$$