12. (10分)已知$a=\sqrt{5}+2,b=\sqrt{5}-2$,求下面的代数式的值:
(1) $a^2 - b^2$;
(2) $a^2 + b^2 + ab$.

12. (1) 因为$a=\sqrt{5}+2,b=\sqrt{5}-2$,所以$a+b=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}$,$a-b=\sqrt{5}+2-\sqrt{5}+2=4$. 所以$a^2-b^2=(a+b)(a-b)=2\sqrt{5}×4=8\sqrt{5}$
(2) 因为$a=\sqrt{5}+2,b=\sqrt{5}-2$,所以$a+b=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}$,$ab=(\sqrt{5}+2)(\sqrt{5}-2)=1$. 所以$a^2+b^2+ab=(a+b)^2-ab=20-1=19$

13. (10分)已知$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}, y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$. 求：
(1) $\frac{1}{x}+\frac{1}{y}$的值；
(2) $\frac{x}{y}+\frac{y}{x}$的值.

13. 由题意可知, $xy = \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} × \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} = 1$, $x = \dfrac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=5+2\sqrt{6}$,$y=\dfrac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=5-2\sqrt{6}$. (1) $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=5+2\sqrt{6}+5-2\sqrt{6}=10$
(2) $\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x^2+y^2}{xy}=\dfrac{(x+y)^2-2xy}{xy}=\dfrac{10^2-2×1}{1}=98$