$解:(2)设分式\frac{a}{2a+1}的“美妙分式”为A$
$则|A-\frac{a}{2a+1}|=3,$
$∴A-\frac{a}{2a+1}=3或A-\frac{a}{2a+1}=-3$
$①当A-\frac{a}{2a+1}=3时,A=\frac{a}{2a+1}+3=\frac{7a+3}{2a+1}$
$②当A-\frac{a}{2a+1}=-3时,A=\frac{a}{2a+1}-3=-\frac{5a+3}{2a+1}$
$综上,分式\frac{a}{2a+1}的“美妙分式”为$
$\frac{7a+3}{2a+1}或-\frac{5a+3}{2a+1}$
$(3)由题,|\frac{4a}{a^{2}-b^{2}}-\frac{a}{a+b}|=3$
$∵|\frac{4a^{2}}{a^{2}-b^{2}}-\frac {a}{a+b}|=|\frac{3a^{2}+ab}{(a+b)(a-b)}|=3$
$∴\frac{3a^{2}+ab}{(a+b)(a-b)}=3或\frac{3a^{2}+ab}{(a+b)(a-b)}=-3$
$∴3a^{2}+ab=3(a^{2}-b^{2})或3a^{2}+ab=-3(a^{2}-b^{2})$
$∵a、b均为不等于0的实数,∴a=-3b或ab=3b^{2}-6a^{2}$
$把a=-3b代入原式=\frac{17b^{2}}{-3b^{2}}=-\frac{17}{3}$
$把ab=3b^{2}-6a^{2}代入原式=\frac{2a^{2}-b^{2}}{-3(2a^{2}-b^{2})}=-\frac{1}{3}$
$综上,分式的值为-\frac{17}{3}或-\frac{1}{3}$