Answer: For every positive integer \(n\), \(3^{3n}-1\) is divisible by \(13\).

We prove the result by induction on \(n\).

Let \(P(n)\) be the statement that \(13\mid (3^{3n}-1)\).

Base case: When \(n=1\),

\(3^{3\cdot 1}-1=3^3-1=27-1=26=13\times 2\),

so \(P(1)\) is true.

Inductive step: Assume that \(P(k)\) is true for some positive integer \(k\). Then

\(3^{3k}-1\) is divisible by \(13\),

so we can write \(3^{3k}-1=13m\) for some integer \(m\).

Now consider \(P(k+1)\):

\(3^{3(k+1)}-1=3^{3k+3}-1=27\cdot 3^{3k}-1\).

Since \(27=26+1\),

\(27\cdot 3^{3k}-1=26\cdot 3^{3k}+(3^{3k}-1)\).

Also \(26\cdot 3^{3k}\) is divisible by \(13\), and by the inductive hypothesis \(3^{3k}-1\) is divisible by \(13\). Therefore their sum is divisible by \(13\).

Hence \(13\mid (3^{3(k+1)}-1)\), so \(P(k+1)\) is true.

Therefore, by mathematical induction, \(3^{3n}-1\) is divisible by \(13\) for every positive integer \(n\).