Four. (20 points) Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms $S_{n}=2 a_{n}-3 \times 2^{n}+4\left(n \in \mathbf{N}_{+}\right)$.
(1) Find the general term formula for the sequence $a_{n}$;
(2) Let $T_{n}$ be the sum of the first $n$ terms of the sequence $\left\{S_{n}-4\right\}$, find $T_{n}$.

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(1) $S_{n+1}-S_{n}=2\left(a_{n+1}-a_{n}\right)-3\left(2^{n+1}-2^{n}\right)$, i.e., $a_{n+1}=2 a_{n}+3 \times 2^{n}$.
Dividing both sides by $2^{n+1}$ gives $\frac{a_{n+1}}{2^{n+1}}=\frac{a_{n}}{2^{n}}+\frac{3}{2}$, thus $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{3}{2}$.
From $a_{1}=2 a_{1}-3 \times 2^{1}+4$, we get $a_{1}=2$.
Therefore, the sequence $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence with the first term $\frac{a_{1}}{2^{1}}=\frac{2}{2}=1$ and common difference $\frac{3}{2}$.
By $\frac{a_{n}}{2^{n}}=1+(n-1) \times \frac{3}{2}$, we have
$$
a_{n}=\left(\frac{3}{2} n-\frac{1}{2}\right) 2^{n} \text {. }
$$
(2) Substituting $a_{n}=\left(\frac{3}{2} n-\frac{1}{2}\right) 2^{n}$ into $S_{n}=2 a_{n}-$ $3 \times 2^{n}+4$, we get
$$
S_{n}-4=(3 n-4) 2^{n} \text {. }
$$

Notice that
$$
\begin{array}{l} 
T_{n}=(3 \times 1-4) \times 2^{1}+(3 \times 2-4) \times 2^{2}+ \\
\cdots+(3 n-4) \times 2^{n}, \\
2 T_{n}=(3 \times 1-4) \times 2^{2}+(3 \times 2-4) \times 2^{3}+ \\
\cdots+(3 n-4) \times 2^{n+1} .
\end{array}
$$
(1) - (2) gives
$$
\begin{aligned}
-T_{n}=(3 \times 1-4) \times 2^{1}+3 \times 2^{2}+3 \times 2^{3}+ \\
\cdots+3 \times 2^{n}-(3 n-4) \times 2^{n+1} \\
=-2+3\left(2^{2}+3^{2}+\cdots+2^{n}\right)-(3 n-4) 2^{n+1} \\
=-2+3 \times \frac{2^{2}\left(1-2^{n-1}\right)}{1-2}-(3 n-4) 2^{n+1} \\
=-14+(14-6 n) 2^{n} .
\end{aligned}
$$

Thus, $T_{n}=14-(14-6 n) 2^{n}$.