Base Case: For \(n = 1\), \(6^{4 \times 1} + 38^1 - 2 = 1296 + 38 - 2 = 1332\). Since \(1332 \div 74 = 18\), it is divisible by 74.

Inductive Step: Assume \(6^{4k} + 38^k - 2\) is divisible by 74 for some positive integer \(k\). This is the inductive hypothesis.

Consider \(6^{4(k+1)} + 38^{k+1} - 2\):

\(6^{4(k+1)} + 38^{k+1} - 2 = 6^{4k+4} + 38^{k+1} - 2\)

\(= (1295 + 1) \times 6^{4k} + (37 + 1) \times 38^k - 2\)

\(= 1295 \times 6^{4k} + 6^{4k} + 37 \times 38^k + 38^k - 2\)

By the inductive hypothesis, \(6^{4k} + 38^k - 2\) is divisible by 74, and since \(1295 \times 6^{4k} + 37 \times 38^k\) is also divisible by 74, the entire expression is divisible by 74.

Therefore, by induction, \(6^{4n} + 38^n - 2\) is divisible by 74 for every positive integer \(n\).