14. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=2, a_{2}+a_{4}=8,
$$

and for any $n \in \mathbf{Z}_{+}$, the function
$$
f(x)=\left(a_{n}-a_{n+1}+a_{n+2}\right) x+a_{n+1} \cos x-a_{n+2} \sin x
$$

satisfies $f^{\prime}\left(\frac{\pi}{2}\right)=0$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If $b_{n}=2\left(a_{n}+\frac{1}{2^{a_{n}}}\right)$, prove:
$$
b_{1}+b_{2}+\cdots+b_{2012}<2013 \times 2014 .
$$
14. (1) From the given, we have
$$
f^{\prime}(x)=a_{n}-a_{n+1}+a_{n+2}-a_{n+1} \sin x-a_{n+2} \cos x \text {. }
$$

Also, \( f^{\prime}\left(\frac{\pi}{2}\right)=0 \), so
$$
a_{n}-a_{n+1}+a_{n+2}-a_{n+1}=0 \text {. }
$$

Thus, \( a_{n+1}-a_{n}=a_{n+2}-a_{n+1} \).
Therefore, \( \left\{a_{n}\right\} \) is an arithmetic sequence.
Given \( a_{1}=2, a_{2}+a_{4}=8 \), the common difference is \( d=1 \).
Hence, the general term of the sequence \( \left\{a_{n}\right\} \) is
$$
a_{n}=2+1 \times(n-1)=n+1 \text {. }
$$
(2) Note that,
$$
\begin{aligned}
b_{n} & =2\left(a_{n}+\frac{1}{2^{a_{n}}}\right)=2\left(n+1+\frac{1}{2^{n+1}}\right)=2 n+\frac{1}{2^{n}}+2 \\
\Rightarrow & b_{1}+b_{2}+\cdots+b_{n} \\
& =2 n+n(n+1)+\frac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}} \\
& =n^{2}+3 n+1-\frac{1}{2^{n}} \\
& <n^{2}+3 n+2=(n+1)(n+2) .
\end{aligned}
$$

Taking \( n=2012 \), we get
$$
b_{1}+b_{2}+\cdots+b_{2012}<2013 \times 2014 .
$$