$[a,a^\dagger]=1$ gives $aa^\dagger=1+a^\dagger a$
So $\{a,a^\dagger\}=aa^\dagger+a^\dagger a=2a^\dagger a+1=2\hat{n}+1$
To calculate the expectation value $\langle 2\hat{n}+1 \rangle$ we have (take $\hbar=1$ )
\begin{align}
\langle 2\hat{n}+1 \rangle&=\text{Tr}[\hat{\rho} (2\hat{n}+1)] \\
&=\text{Tr}[\frac{e^{-\beta \omega (\hat{n}+1/2)}}{\text{Tr}[e^{-\beta \omega (\hat{n}+1/2)}]} (2\hat{n}+1)] \\
\end{align}
First let's compute $Z=\text{Tr}[e^{-\beta \omega (\hat{n}+1/2)}]=e^{-\beta \omega/2}\sum_{n}\langle n|e^{-\beta \omega \hat{n}}|n \rangle=e^{-\beta \omega/2}\sum_{n}e^{-\beta \omega n} $
Then
\begin{align}
\text{Tr}[\frac{e^{-\beta \omega (\hat{n}+1/2)}}{Z} (2\hat{n}+1)]&=1+2\frac{e^{-\beta \omega/2} \sum_{n}n e^{-n \beta \omega}}{Z} \\
&=1+2\frac{ \sum_{n}n e^{-n \beta \omega}}{\sum_{n}e^{-\beta \omega n}} \\
\langle 2\hat{n}+1 \rangle&=1+\frac{2}{e^{\beta \omega}-1}
\end{align}
Hope this is helpful.