If $60^a = 3$ and $60^b = 5$, then $12^{[(1-a-b)/2(1-b)]}$ is

$\text{(A)} \ \sqrt{3} \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \sqrt{5} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \sqrt{12}$
1. Given the equations \(60^a = 3\) and \(60^b = 5\), we can express \(a\) and \(b\) in terms of logarithms:
   \[
   a = \frac{\log 3}{\log 60}
   \]
   \[
   b = \frac{\log 5}{\log 60}
   \]

2. We need to find the value of \(12^{\left(\frac{1-a-b}{2(1-b)}\right)}\). First, let's simplify the exponent \(\frac{1-a-b}{2(1-b)}\):
   \[
   1 - a - b = 1 - \frac{\log 3}{\log 60} - \frac{\log 5}{\log 60} = 1 - \frac{\log 3 + \log 5}{\log 60}
   \]
   Using the property of logarithms \(\log(ab) = \log a + \log b\), we get:
   \[
   1 - a - b = 1 - \frac{\log(3 \cdot 5)}{\log 60} = 1 - \frac{\log 15}{\log 60}
   \]

3. Next, we simplify \(2(1 - b)\):
   \[
   2(1 - b) = 2 \left(1 - \frac{\log 5}{\log 60}\right) = 2 \left(\frac{\log 60 - \log 5}{\log 60}\right) = 2 \left(\frac{\log \left(\frac{60}{5}\right)}{\log 60}\right) = 2 \left(\frac{\log 12}{\log 60}\right)
   \]

4. Now, we combine the results:
   \[
   \frac{1 - a - b}{2(1 - b)} = \frac{1 - \frac{\log 15}{\log 60}}{2 \left(\frac{\log 12}{\log 60}\right)} = \frac{\frac{\log 60 - \log 15}{\log 60}}{2 \left(\frac{\log 12}{\log 60}\right)} = \frac{\log 60 - \log 15}{2 \log 12}
   \]
   Simplifying further using the property \(\log(a) - \log(b) = \log\left(\frac{a}{b}\right)\):
   \[
   \frac{\log \left(\frac{60}{15}\right)}{2 \log 12} = \frac{\log 4}{2 \log 12}
   \]

5. We know that \(\log 4 = 2 \log 2\), so:
   \[
   \frac{2 \log 2}{2 \log 12} = \frac{\log 2}{\log 12}
   \]

6. Therefore, the exponent simplifies to:
   \[
   12^{\left(\frac{\log 2}{\log 12}\right)} = 2
   \]

The final answer is \(\boxed{2}\).