解:因为$\vert a - 4\vert + b^{2}+\frac{1}{4}=b,$所以$\vert a - 4\vert+(b - \frac{1}{2})^{2}=0。$
因为$\vert a - 4\vert\geq0,$$(b - \frac{1}{2})^{2}\geq0,$所以$a = 4,$$b = \frac{1}{2}。$
又因为原式$=\frac{a(a^{2}-4b^{2})}{a(a^{2}-4ab + 4b^{2})}=\frac{a + 2b}{a - 2b},$
把$a = 4,$$b = \frac{1}{2}$代入,得原式$=\frac{4 + 2\times\frac{1}{2}}{4 - 2\times\frac{1}{2}}=\frac{4 + 1}{4 - 1}=\frac{5}{3}。$