解:(1)、
$\begin{align*}\sum_{n=1}^{\infty }\frac{1}{n}(a_n+a_{n+2})&=\sum_{n=1}^{\infty }\frac{1}{n}\int_{0}^{\frac{\pi}{4}}\tan^nx(1+\tan^2x)dx\\\\&=\sum_{n=1}^{\infty }\frac{1}{n}\tan^nxd\tan x\\\\&=\sum_{n=1}^{\infty }\frac{1}{n}\cdot \frac{1}{n+1}\tan^{n+1}x|_0^{\frac{\pi}{4}}\\\\&=\sum_{n=1}^{\infty }\frac{1}{n(n+1)}\\\\&=\lim_{n \to \infty }\sum_{k=1}^{n}(\frac{1}{k}-\frac{1}{k+1})\\\\&=\lim_{n \to \infty }(1-\frac{1}{n+1})\\\\&=1.
\end{align*}$
(2)、
$\displaystyle \because \frac{a_n}{n^\lambda }=\frac{1}{n^\lambda }\int_{0}^{\frac{\pi}{4}}\tan^nxdx\leq \frac{1}{n^\lambda }\int_{0}^{\frac{\pi}{4}}\tan^{n-2}xd\tan x=\frac{1}{n^\lambda (n-1)}< \frac{1}{n^{\lambda +1}}.$
$\displaystyle \therefore \lambda > 0,\sum_{n=1}^{\infty }\frac{a_n}{n^\lambda }< \infty .$