解：由题意得$M_{1}= \sqrt {x+1}+ \sqrt {x}$，$N_{1}= \sqrt {x+1}- \sqrt x$
$M_{2} =2 \sqrt {x+1}$，$N_{2}= 2 \sqrt {x}$，
$M_{3} =2 \sqrt {x+1}+2 \sqrt x$，$N_{2}=2 \sqrt {x+1}-2\sqrt x$，$M_{4}=4 \sqrt {x+1}$，$N=4\sqrt x······$
∴$M_{2n+1} =2^n \sqrt {x+1}+2^n \sqrt {x}$，$N_{2n+1}=2^n \sqrt {x+1}-2^n\sqrt x$，
$M_{2n+2}=2^{n+1}\sqrt {x+1}$，$N_{2n+2}=2^{n+1}\sqrt x(n$是自然数$).$
$(1)$∵$N_{2n+2}=2^{n+1} \sqrt {x}(n$是自然数$)$
∴$N_{2}=2 \sqrt {x}$，$N_{4}=4 \sqrt {x}$，$N_{6}=8 \sqrt x$，$N_{8}=16\sqrt x$
即$N_{2}+N_{4}+N_{6}+N_{8}=2 \sqrt x+4 \sqrt x+8 \sqrt {x}+16 \sqrt {x}=30 \sqrt {x}$
∵又$x=1$，∴$N_{2}+N_{4}+N_{6}+N_{8}=30$
$(2)$∵$M_{2n+2}=2^{n+1}\sqrt {x+1}(n$是自然数$)$，∴$M_{12}=2^6 \sqrt {x+1}=64 \sqrt {x+1}$
$(3)$∵$M_{2n+1}=2^n\sqrt {x+1}+2^n \sqrt {x}$，$N_{2n+1}= 2^n \sqrt {x+1}-2^n \sqrt {x}$
∴$M_{2n+1} · N_{2n+1}=(2^n \sqrt {x+1}+2^n\sqrt {x}) · (2^n \sqrt {x+1}-2^n\sqrt x)$
$=2^{2n}[(\sqrt {x+1})²-(\sqrt {x})²]=2^{2n}(x+1-x)=2^{2n}$
