1. Given a sequence of non-negative numbers $\left\{a_{n}\right\}$ with the first term being $a$, the sum of the first $n$ terms is $S_{n}$, and $S_{n}=\left(\sqrt{S_{n-1}}+\sqrt{a}\right)^{2}$. If $b_{n}=$ $\frac{a_{n+1}}{a_{n}}+\frac{a_{n}}{a_{n+1}}$, the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $T_{n}$, then $T_{n}=$
$$
=1 \cdot \frac{4 n^{2}+6 n}{2 n+1} \text {. }
$$

From $S_{n}=\left(\sqrt{S_{n-1}}+\sqrt{a}\right)^{2}$, we get
$$
\sqrt{S_{n}}=\sqrt{S_{n-1}}+\sqrt{a} \text {. }
$$

Also, $\sqrt{S_{1}}=\sqrt{a_{1}}=\sqrt{a}$, then $\left\{\sqrt{S_{n}} \mid\right.$ is an arithmetic sequence with the first term $\sqrt{a}$ and common difference $\sqrt{a}$, that is
$$
\sqrt{S_{n}}=\sqrt{a}+(n-1) \sqrt{a}=n \sqrt{a} \text {. }
$$

Thus, $S_{n}=a n^{2}$.
So $a_{n}=S_{n}-S_{n-1}=a n^{2}-a(n-1)^{2}=2 a n-a$,
$$
\begin{array}{l}
b_{n}=\frac{a_{n+1}}{a_{n}}+\frac{a_{n}}{a_{n+1}} \\
=\frac{2 a(n+1)-a}{2 a n-a}+\frac{2 a n-a}{2 a(n+1)-a} \\
=\frac{2 n+1}{2 n-1}+\frac{2 n-1}{2 n+1}=2+2\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right) .
\end{array}
$$

Then $T_{n}=b_{1}+b_{2}+\cdots+b_{n}$
$$
=2 n+2\left(1-\frac{1}{2 n+1}\right)=\frac{4 n^{2}+6 n}{2 n+1} \text {. }
$$