[b]Q11.[/b] Let be given a sequense $a_1=5, \; a_2=8$ and $a_{n+1}=a_n+3a_{n-1}, \qquad n=1,2,3,...$ Calculate the greatest common divisor of $a_{2011}$ and $a_{2012}$.
1. **Initial Terms and Recurrence Relation:**
   Given the sequence:
   \[
   a_1 = 5, \quad a_2 = 8
   \]
   and the recurrence relation:
   \[
   a_{n+1} = a_n + 3a_{n-1} \quad \text{for} \quad n = 1, 2, 3, \ldots
   \]

2. **Modulo 3 Analysis:**
   We start by examining the sequence modulo 3:
   \[
   a_1 \equiv 5 \equiv 2 \pmod{3}
   \]
   \[
   a_2 \equiv 8 \equiv 2 \pmod{3}
   \]
   Using the recurrence relation modulo 3:
   \[
   a_{n+1} \equiv a_n + 3a_{n-1} \equiv a_n \pmod{3}
   \]
   This implies:
   \[
   a_{n+1} \equiv a_n \pmod{3}
   \]
   Since \(a_1 \equiv 2 \pmod{3}\) and \(a_2 \equiv 2 \pmod{3}\), it follows that:
   \[
   a_n \equiv 2 \pmod{3} \quad \text{for all} \quad n \geq 1
   \]

3. **Greatest Common Divisor Analysis:**
   We need to find \(\gcd(a_{2011}, a_{2012})\). From the recurrence relation:
   \[
   a_{2012} = a_{2011} + 3a_{2010}
   \]
   Therefore:
   \[
   \gcd(a_{2011}, a_{2012}) = \gcd(a_{2011}, a_{2011} + 3a_{2010})
   \]
   Using the property of gcd:
   \[
   \gcd(a, b) = \gcd(a, b - ka) \quad \text{for any integer} \quad k
   \]
   We get:
   \[
   \gcd(a_{2011}, a_{2011} + 3a_{2010}) = \gcd(a_{2011}, 3a_{2010})
   \]
   Since \(a_{2011}\) and \(a_{2010}\) are both congruent to 2 modulo 3, we have:
   \[
   \gcd(a_{2011}, 3a_{2010}) = \gcd(a_{2011}, 3)
   \]
   Because \(a_{2011} \equiv 2 \pmod{3}\), it is not divisible by 3. Therefore:
   \[
   \gcd(a_{2011}, 3) = 1
   \]

Conclusion:
\[
\boxed{1}
\]