解$：$
任务一:
$\frac {1}{\sqrt {2}-1}=\frac {\sqrt {2}+1}{(\sqrt {2}-1)(\sqrt {2}+1)}=\frac {\sqrt {2}+1}{2 - 1}=\sqrt {2}+1$
任务二:由 $MN$ 为 $x$,知 $BC = MB = x$, $AC=\frac {1}{2}x$,
$∴AB=\sqrt {BC^2+AC^2}=\sqrt {x^2+(\frac {1}{2}x)^2}=\sqrt {x^2+\frac {x^2}{4}}=\frac {\sqrt {5}}{2}x$
证明:由折叠的性质可以知道, $AD = AB$,
$∴CD = AD - AC = AB - AC=\frac {\sqrt {5}}{2}x-\frac {1}{2}x=\frac {\sqrt {5}-1}{2}x$
$∴CD：BC=\frac {\sqrt {5}-1}{2}x：x=\frac {\sqrt {5}-1}{2}$
$∴$ 矩形 $BCDE$ 是黄金矩形.
任务三:如答图,连接 $CE$,过点 $E$ 作 $EH\perp MC$ 于点 $H$.
由图知 $S_{\triangle MCE}=S_{\triangle MBC}+S_{\triangle BCE}$,
$S_{\triangle MCE}=\frac {1}{2}MB·BC+\frac {1}{2}BC·BE=\frac {1}{2}x^2+\frac {1}{2}x·\frac {\sqrt {5}-1}{2}x=\frac {1}{2}x^2·\frac {\sqrt {5}+1}{2}$
$∵MN = 2，∴S_{\triangle MCE}=\frac {1}{2}×2^2×\frac {\sqrt {5}+1}{2}=\sqrt {5}+1$
$MC=\sqrt {MB^2+BC^2}=2\sqrt {2}$
$∴S_{\triangle MCE}=\frac {1}{2}×2\sqrt {2}·EH=\sqrt {2}EH=\sqrt {5}+1$
$∴EH=\frac {\sqrt {5}+1}{\sqrt {2}}=\frac {\sqrt {10}+\sqrt {2}}{2}$
$∴$ 点 $E$ 到线段 $MC$ 的距离是 $\frac {\sqrt {10}+\sqrt {2}}{2}$.
