14. Given that $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, and $4 S_{n}=3 a_{n}+2^{n+1}(n \in \mathbf{N})$.
Find: (1) the relationship between $a_{n}$ and $a_{n-1}$;
(2) all values of $a_{0}$ that make the sequence $a_{0}, a_{1}, \cdots$ increasing.
14. (1) From the problem, we have
$$
4 S_{n-1}=3 a_{n-1}+2^{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Subtracting the given equation from the above equation, we get
$$
\begin{array}{l}
4 a_{n}=3 a_{n}-3 a_{n-1}+2^{n} \\
\Rightarrow a_{n}=2^{n}-3 a_{n-1}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
(2) From (1), dividing both sides by $2^{n}$, we get
$$
\begin{aligned}
\frac{a_{n}}{2^{n}} & =-\frac{3}{2} \times \frac{a_{n-1}}{2^{n-1}}+1 \\
\Rightarrow & \frac{a_{n}}{2^{n}}-\frac{2}{5}=-\frac{3}{2}\left(\frac{a_{n-1}}{2^{n-1}}-\frac{2}{5}\right) \\
\Rightarrow & \frac{a_{n}}{2^{n}}-\frac{2}{5}=\left(-\frac{3}{2}\right)^{n}\left(\frac{a_{0}}{2^{0}}-\frac{2}{5}\right) \\
\Rightarrow & a_{n}=2^{n}\left[\frac{2}{5}+\left(-\frac{3}{2}\right)^{n}\left(a_{0}-\frac{2}{5}\right)\right] \\
& =(-3)^{n}\left(a_{0}-\frac{2}{5}\right)+2^{n} \times \frac{2}{5} .
\end{aligned}
$$

Thus, $a_{n+1}-a_{n}$
$$
\begin{array}{l}
=\left(a_{0}-\frac{2}{5}\right)\left[(-3)^{n+1}-(-3)^{n}\right]+\frac{2}{5}\left(2^{n+1}-2^{n}\right) \\
=(-3)^{n}\left(a_{0}-\frac{2}{5}\right) \times(-4)+\frac{2}{5} \times 2^{n} .
\end{array}
$$

If $a_{0}>\frac{2}{5}$, then for sufficiently large even $n$,
$$
a_{n+1}<a_{n} ;
$$

If $a_{0}<\frac{2}{5}$, then for sufficiently large odd $n$,
$$
a_{n+1}<a_{n} \text {. }
$$

In summary, $a_{0}=\frac{2}{5}$.