12. Let $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, and $a_{n+1}=2 S_{n}+1\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Insert $n$ real numbers between $a_{n}$ and $a_{n+1}$ such that these $n+2$ numbers form an arithmetic sequence with a common difference of $d_{n}$. Let the sum of the first $n$ terms of the sequence $\left\{\frac{1}{d_{n}}\right\}$ be $T_{n}$. Prove that: $T_{n}<\frac{15}{8}$.
(1) From $\left\{\begin{array}{l}a_{n+1}=2 S_{n}+1 \\ a_{n}=2 S_{n-1}+1(n \geq 2)\end{array}\right.$, subtracting the two equations gives
$$
a_{n+1}-a_{n}=2\left(S_{n}-S_{n-1}\right)=2 a_{n}
$$

Thus, $a_{n+1}=3 a_{n}(n \geq 2)$.
Since $\left\{a_{n}\right\}$ is a geometric sequence and $a_{2}=2 a_{1}+1$, we have $3 a_{1}=2 a_{1}+1$, so $a_{1}=1$, and thus $a_{n}=3^{n-1}$;
(2) From the given, we have $a_{n+1}=a_{n}+(n+1) d_{n}$, so
$$
\frac{1}{d_{n}}=\frac{n+1}{a_{n+1}-a_{n}}=\frac{n+1}{2 \cdot 3^{n-1}}
$$

Therefore,
$$
2 T_{n}=2+\frac{3}{3}+\frac{4}{3^{2}}+\cdots+\frac{n+1}{3^{n-1}}
$$

Then,
$$
\frac{2}{3} T_{n}=\frac{2}{3}+\frac{3}{3^{2}}+\cdots+\frac{n}{3^{n-1}}+\frac{n+1}{3^{n}}
$$

Subtracting the two equations gives:
$$
\frac{4}{3} T_{n}=2+\frac{1}{3}+\frac{1}{3^{2}}+\cdots+\frac{1}{3^{n-1}}-\frac{n+1}{3^{n}}=2+\frac{\frac{1}{3}\left(1-\frac{1}{3^{n-1}}\right)}{1-\frac{1}{3}}-\frac{n+1}{3^{n}}
$$

Thus,
$$
T_{n}=\frac{15}{8}-\frac{2 n+5}{8 \cdot 3^{n-1}}<\frac{15}{8}
$$