$解:(1) \frac{2}{\sqrt {3} -1} = \frac{2(\sqrt{3}+1)}{\sqrt{3}+1)(\sqrt{3}-1)}\ = \sqrt{3} +1.$
$(2)原式= \frac{1}{2} ( \sqrt{3} -1+ \sqrt{5} - \sqrt{3} + \sqrt{7}$
$\ -\sqrt{5} +···+ \sqrt{2023} - \sqrt{2021} )\ $
$= \frac{1}{2} ( \sqrt{2023} -1)= \frac{\sqrt{2023}-1}{2}\ $
$(3)∵a= \frac{1}{\sqrt {2} -1} = \frac{\sqrt{2}+1}{(\sqrt {2} -1)\sqrt{2}+1)} = \sqrt{2} +1，\ $
$∴a-1= \sqrt{2} ，\ $
$∴a²-2a+1=2，\ $
$∴a²-2a=1.\ $
$① \frac{1}{2} a²-a-1= \frac{1}{2} (a²-2a)-1\ $
$= \frac{1}{2} ×1-1=- \frac{1}{2} .\ $
$②2a³-5a²+1=2a(a²-2a)-a²+1\ $
$=2a-a²+1=-1+1=0.\ $
