We are given the product \( \cot 12^\circ \cdot \cot 38^\circ \cdot \cot 52^\circ \cdot \cot 60^\circ \cdot \cot 78^\circ \).
Using the identity \( \cot \theta = \tan (90^\circ - \theta) \), we can simplify the angles:
\[
\cot 12^\circ = \tan 78^\circ, \quad \cot 38^\circ = \tan 52^\circ.
\]
Thus, the product becomes:
\[
\tan 78^\circ \cdot \tan 52^\circ \cdot \cot 60^\circ = \tan 78^\circ \cdot \tan 52^\circ \cdot \frac{1}{\tan 30^\circ}.
\]
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), the product simplifies to:
\[
\tan 78^\circ \cdot \tan 52^\circ \cdot \sqrt{3} = 1.
\]
Therefore, the answer is \( \boxed{1} \).