解：(1)把$M(2,m)$代入$y = x$得$m = 2，$把$M(2,2)$代入$y = -\dfrac{1}{2}x + b$得$2=-\dfrac{1}{2}\times2 + b，$$2=-1 + b，$解得$b = 3。$
對(duì)于$y = -\dfrac{1}{2}x + 3，$令$y = 0，$則$0=-\dfrac{1}{2}x + 3，$$\dfrac{1}{2}x = 3，$$x = 6，$所以$A(6,0)。$
(2)對(duì)于$y = -\dfrac{1}{2}x + 3，$令$x = 0，$則$y = 3，$所以$B(0,3)，$$OB = 3。$
$C\left(a,-\dfrac{1}{2}a + 3\right)，$$D(a,a)，$$CD=a-\left(-\dfrac{1}{2}a + 3\right)=a+\dfrac{1}{2}a - 3=\dfrac{3}{2}a - 3。$
因?yàn)?OB = CD，$所以$\dfrac{3}{2}a - 3=3，$$\dfrac{3}{2}a = 6，$解得$a = 4。$
(3) $OA = 6，$$OD=\sqrt{a^{2}+a^{2}}=\sqrt{2}a，$$AD=\sqrt{(a - 6)^{2}+a^{2}}。$
當(dāng)$OA = OD$時(shí)，$\sqrt{2}a=6，$解得$a = 3\sqrt{2}；$
當(dāng)$OA = AD$時(shí)，$(a - 6)^{2}+a^{2}=36，$$a^{2}-12a + 36+a^{2}=36，$$2a^{2}-12a=0，$$a^{2}-6a=0，$$a(a - 6)=0，$因?yàn)?a＞2，$所以$a = 6；$
當(dāng)$OD = AD$時(shí)，$a^{2}+a^{2}=(a - 6)^{2}+a^{2}，$$a^{2}=a^{2}-12a + 36，$$12a = 36，$解得$a = 3。$
綜上，$a$的值為$3\sqrt{2}$或$3$或$6。$
