Let $a$ be a real  number such that $\left(a + \frac{1}{a}\right)^2=11$. What possible values can $a^3 + \frac{1}{a^3}$ and $a^5 + \frac{1}{a^5}$ take?
1. Given the equation:
   \[
   \left(a + \frac{1}{a}\right)^2 = 11
   \]
   Expanding and simplifying, we get:
   \[
   a^2 + 2 + \frac{1}{a^2} = 11
   \]
   Subtracting 2 from both sides:
   \[
   a^2 + \frac{1}{a^2} = 9 \quad \text{(Equation II)}
   \]

2. From the given equation, we have:
   \[
   a + \frac{1}{a} = \sqrt{11} \quad \text{or} \quad a + \frac{1}{a} = -\sqrt{11} \quad \text{(Equation I)}
   \]

3. To find \(a^3 + \frac{1}{a^3}\), we use the identity:
   \[
   (a + \frac{1}{a})(a^2 + \frac{1}{a^2}) = a^3 + \frac{1}{a^3} + a + \frac{1}{a}
   \]
   Substituting \(a + \frac{1}{a}\) and \(a^2 + \frac{1}{a^2}\) from Equations I and II:
   \[
   (a + \frac{1}{a})(a^2 + \frac{1}{a^2}) = \sqrt{11} \cdot 9 = 9\sqrt{11}
   \]
   \[
   a^3 + \frac{1}{a^3} + a + \frac{1}{a} = 9\sqrt{11}
   \]
   \[
   a^3 + \frac{1}{a^3} = 9\sqrt{11} - \sqrt{11} = 8\sqrt{11}
   \]
   Similarly, for \(a + \frac{1}{a} = -\sqrt{11}\):
   \[
   a^3 + \frac{1}{a^3} = 9(-\sqrt{11}) - (-\sqrt{11}) = -8\sqrt{11}
   \]

4. To find \(a^5 + \frac{1}{a^5}\), we use the identity:
   \[
   (a^2 + \frac{1}{a^2})^2 = a^4 + \frac{1}{a^4} + 2
   \]
   Substituting \(a^2 + \frac{1}{a^2} = 9\):
   \[
   81 = a^4 + \frac{1}{a^4} + 2
   \]
   \[
   a^4 + \frac{1}{a^4} = 79
   \]

5. Using the identity:
   \[
   (a^4 + \frac{1}{a^4})(a + \frac{1}{a}) = a^5 + \frac{1}{a^5} + a^3 + \frac{1}{a^3}
   \]
   Substituting \(a^4 + \frac{1}{a^4}\) and \(a + \frac{1}{a}\):
   \[
   79\sqrt{11} = a^5 + \frac{1}{a^5} + 8\sqrt{11}
   \]
   \[
   a^5 + \frac{1}{a^5} = 79\sqrt{11} - 8\sqrt{11} = 71\sqrt{11}
   \]
   Similarly, for \(a + \frac{1}{a} = -\sqrt{11}\):
   \[
   79(-\sqrt{11}) = a^5 + \frac{1}{a^5} - 8\sqrt{11}
   \]
   \[
   a^5 + \frac{1}{a^5} = -79\sqrt{11} + 8\sqrt{11} = -71\sqrt{11}
   \]

Conclusion:
\[
(a^3 + \frac{1}{a^3}, a^5 + \frac{1}{a^5}) = (8\sqrt{11}, 71\sqrt{11}), (-8\sqrt{11}, -71\sqrt{11})
\]

The final answer is \( \boxed{ (8\sqrt{11}, 71\sqrt{11}), (-8\sqrt{11}, -71\sqrt{11}) } \)