$解:(1)①\sqrt{4-2\sqrt {3}}= \sqrt{3-2\sqrt {3}+1}$
$= \sqrt{(\sqrt {3})^{2}-2×3×1+1^{2}}=\sqrt{(\sqrt {3}-1)^{2}}=\sqrt {3}-1$
$② \sqrt{7-4\sqrt {7}}=\sqrt{(2-\sqrt{3})^{2}}=2-\sqrt{3}$
$(2)∵a+6\sqrt{5}=(m+\sqrt{5}n)^{2}=m^{2}+5n^{2}+2\sqrt{5}mn$
$∴a=m^{2}+5n^{2}且 2\sqrt {5}mn=6\sqrt {5},∴mn=3$
$∵a、m、n為正整數(shù)$
$∴當(dāng)m=1，n=3時(shí)，a=46;$
$當(dāng)m=3，n=1時(shí)，a=14$
$∴a的值為14或46$
