jmc

We have that $a={\log }_{60}3$ and $b={\log }_{60}5,$ so $$1-a-b={\log }_{60}60-{\log }_{60}3-{\log }_{60}5={\log }_{60}\frac{60}{3\cdot 5}={\log }_{60}4=2{\log }_{60}2$$and $$2(1-b)=2({\log }_{60}60-{\log }_{60}5)=2{\log }_{60}12,$$so $$\frac{1-a-b}{2(1-b)}=\frac{2{\log }_{60}2}{2{\log }_{60}12}={\log }_{12}2.$$Therefore, $$12^{(1-a-b)/(2(1-b))}=\boxed{2}.$$