(a) Base case: For \(n = 1\), \(u_1 = 5 = 6^1 - 1\). Thus, the base case holds.

Inductive step: Assume \(u_k = 6^k - 1\) is true for \(n = k\). Then,

\(u_{k+1} = 6u_k + 5 = 6(6^k - 1) + 5 = 6^{k+1} - 6 + 5 = 6^{k+1} - 1.\)

Thus, \(u_{k+1} = 6^{k+1} - 1\) is true. By induction, \(u_n = 6^n - 1\) for all positive integers \(n\).

(b) We have \(u_{2n} = 6^{2n} - 1\) and \(u_n = 6^n - 1\). Consider:

\(\frac{u_{2n}}{u_n} = \frac{6^{2n} - 1}{6^n - 1} = \frac{(6^n - 1)(6^n + 1)}{6^n - 1} = 6^n + 1.\)

Since \(6^n + 1\) is an integer, \(u_{2n}\) is divisible by \(u_n\) for \(n \geq 1\).