$解：(1) 原式 ={(x+y)}^{2}$
$將x=\sqrt {3}+1，y=\sqrt {3}-1代入，得$
$原式 =\left [ {{({\sqrt {3}+1})+({\sqrt {3}-1})}} \right ]^{2}$
$={(2{\sqrt {3}})}^{2}$
$=12$
$(2) 原式 =(x+y)(x-y)$
$將x=\sqrt {3}+1，y=\sqrt {3}-1代入，得$
$原式 =\left [ {{({\sqrt {3}+1})+({\sqrt {3}}-1)}} \right ]×\left [ {{({\sqrt {3}+1})-({\sqrt {3}-1})}} \right ]$
$=2{\sqrt {3}}×2$
$=4{\sqrt {3}}$
