解:$(1) $原式$=\frac {\sqrt {3}-1}{(\sqrt {3}+1)×(\sqrt {3}-1)}+\frac {\sqrt {5}-\sqrt {3}}{(\sqrt {5}+\sqrt {3})×(\sqrt {5}-\sqrt {3})}$
$+\frac {\sqrt {7}-\sqrt {5}}{(\sqrt {7}+\sqrt {5})×(\sqrt {7}-\sqrt {5})}+…+\frac {\sqrt {121}-\sqrt {119}}{(\sqrt {121}+\sqrt {119})×(\sqrt {121}-\sqrt {119})}$
$=\frac {\sqrt {3}-1}{2}+\frac {\sqrt {5}-\sqrt {3}}{2}+\frac {\sqrt {7}-\sqrt {5}}{2}+…+\frac {\sqrt {121}-\sqrt {119}}{2}$
$=\frac {1}{2}×(\sqrt {3}-1+\sqrt {5}-\sqrt {3}+\sqrt {7}-\sqrt {5}+…+\sqrt {121}-\sqrt {119})$
$=\frac {1}{2}×(\sqrt {121}-1)=5 $
$(2) ① $∵$a=\frac {1}{\sqrt {2}-1}=\frac {\sqrt {2}+1}{(\sqrt {2}-1)×(\sqrt {2}+1)}=\sqrt {2}+1,$
∴$a-1=\sqrt {2}.$∴$(a-1)^2=2.$∴$a^2-2a+1=2.$∴$a^2-2a=1.$
∴$3a^2-6a+1=3(a^2-2a)+1=3×1+1=4 $