14. Let the function
$$
f(x)=\left\{\begin{array}{ll}
(3-a) x-3, & x \leqslant 7 ; \\
a^{x-6} & x>7,
\end{array}\right.
$$

The sequence $\left\{a_{n}\right\}$ satisfies $a_{n}=f(n)\left(n \in \mathbf{N}_{+}\right)$, and the sequence $\left\{a_{n}\right\}$ is an increasing sequence. Then the range of the real number $a$ is $\qquad$ .
14. $(2,3)$.

Since the sequence $\left\{a_{n}\right\}$ is an increasing sequence, we have,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a-3>0, \\
a>1
\end{array} \Rightarrow 12 .
\end{array}
$$

Therefore, the range of the real number $a$ is $(2,3)$.