Define sequences $\{a_n\},\ \{b_n\}$ by 
\[a_n=\int_{-\frac {\pi}6}^{\frac{\pi}6} e^{n\sin \theta}d\theta,\  b_n=\int_{-\frac {\pi}6}^{\frac{\pi}6} e^{n\sin \theta}\cos \theta d\theta\ (n=1,\ 2,\ 3,\ \cdots).\]

(1) Find $b_n$.

(2) Prove that for each $n$, $b_n\leq a_n\leq \frac 2{\sqrt{3}}b_n.$

(3) Find $\lim_{n\to\infty} \frac 1{n}\ln (na_n).$
1. **Finding \( b_n \):**

   We start with the given definition of \( b_n \):
   \[
   b_n = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \cos \theta \, d\theta
   \]
   Using the substitution \( u = \sin \theta \), we have \( du = \cos \theta \, d\theta \). The limits of integration change accordingly: when \( \theta = -\frac{\pi}{6} \), \( u = -\frac{1}{2} \); and when \( \theta = \frac{\pi}{6} \), \( u = \frac{1}{2} \). Thus, the integral becomes:
   \[
   b_n = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{nu} \, du
   \]
   Evaluating this integral, we get:
   \[
   b_n = \left[ \frac{1}{n} e^{nu} \right]_{-\frac{1}{2}}^{\frac{1}{2}} = \frac{1}{n} \left( e^{n/2} - e^{-n/2} \right) = \frac{2}{n} \sinh \frac{n}{2}
   \]

2. **Proving \( b_n \leq a_n \leq \frac{2}{\sqrt{3}} b_n \):**

   We start with the given definitions:
   \[
   a_n = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \, d\theta
   \]
   \[
   b_n = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \cos \theta \, d\theta
   \]
   Note that for all \( \theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \), we have:
   \[
   \cos \frac{\pi}{6} \leq \cos \theta \leq 1
   \]
   Therefore:
   \[
   \cos \theta \leq 1 \leq \frac{\cos \theta}{\cos \frac{\pi}{6}}
   \]
   This implies:
   \[
   e^{n \sin \theta} \cos \theta \leq e^{n \sin \theta} \leq \frac{e^{n \sin \theta} \cos \theta}{\cos \frac{\pi}{6}}
   \]
   Integrating both sides over the interval \(\left[-\frac{\pi}{6}, \frac{\pi}{6}\right]\), we get:
   \[
   \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \cos \theta \, d\theta \leq \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \, d\theta \leq \frac{1}{\cos \frac{\pi}{6}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \cos \theta \, d\theta
   \]
   Simplifying, we get:
   \[
   b_n \leq a_n \leq \frac{2}{\sqrt{3}} b_n
   \]

3. **Finding \( \lim_{n \to \infty} \frac{1}{n} \ln (n a_n) \):**

   Using the results from part 2, we have:
   \[
   b_n \leq a_n \leq \frac{2}{\sqrt{3}} b_n
   \]
   Taking the natural logarithm and dividing by \( n \), we get:
   \[
   \frac{1}{n} \ln b_n \leq \frac{1}{n} \ln a_n \leq \frac{1}{n} \ln \left( \frac{2}{\sqrt{3}} b_n \right)
   \]
   Since \( b_n = \frac{2}{n} \sinh \frac{n}{2} \), we have:
   \[
   \ln b_n = \ln \left( \frac{2}{n} \sinh \frac{n}{2} \right) = \ln 2 - \ln n + \ln \sinh \frac{n}{2}
   \]
   For large \( n \), \( \sinh \frac{n}{2} \approx \frac{1}{2} e^{n/2} \), so:
   \[
   \ln \sinh \frac{n}{2} \approx \ln \left( \frac{1}{2} e^{n/2} \right) = \frac{n}{2} - \ln 2
   \]
   Thus:
   \[
   \ln b_n \approx \ln 2 - \ln n + \frac{n}{2} - \ln 2 = \frac{n}{2} - \ln n
   \]
   Therefore:
   \[
   \frac{1}{n} \ln b_n \approx \frac{1}{n} \left( \frac{n}{2} - \ln n \right) = \frac{1}{2} - \frac{\ln n}{n}
   \]
   As \( n \to \infty \), \( \frac{\ln n}{n} \to 0 \), so:
   \[
   \lim_{n \to \infty} \frac{1}{n} \ln b_n = \frac{1}{2}
   \]
   Similarly:
   \[
   \frac{1}{n} \ln \left( \frac{2}{\sqrt{3}} b_n \right) = \frac{1}{n} \left( \ln \frac{2}{\sqrt{3}} + \ln b_n \right) \approx \frac{1}{n} \ln \frac{2}{\sqrt{3}} + \frac{1}{2} - \frac{\ln n}{n}
   \]
   As \( n \to \infty \), \( \frac{1}{n} \ln \frac{2}{\sqrt{3}} \to 0 \) and \( \frac{\ln n}{n} \to 0 \), so:
   \[
   \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{2}{\sqrt{3}} b_n \right) = \frac{1}{2}
   \]
   By the squeeze theorem:
   \[
   \lim_{n \to \infty} \frac{1}{n} \ln a_n = \frac{1}{2}
   \]
   Therefore:
   \[
   \lim_{n \to \infty} \frac{1}{n} \ln (n a_n) = \lim_{n \to \infty} \left( \frac{1}{n} \ln n + \frac{1}{n} \ln a_n \right) = 0 + \frac{1}{2} = \frac{1}{2}
   \]

The final answer is \( \boxed{ \frac{1}{2} } \).