3.3.3 ** Given a monotonically increasing geometric sequence $\left\{a_{n}\right\}$ that satisfies $a_{2}+a_{3}+a_{4}=28$, and $a_{3}+2$ is the arithmetic mean of $a_{2}$ and $a_{4}$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If $b_{n}=a_{n} \log _{\frac{1}{2}} a_{n}, S_{n}=b_{1}+b_{2}+\cdots+b_{n}$, find the smallest positive integer $n$ such that $S_{n}+n \cdot 2^{n+1}>50$.
(1) Let the common ratio of the geometric sequence be $q(\neq 0)$, then from the given conditions we have
$$
\begin{array}{l}
a_{1} q+a_{1} q^{2}+a_{1} q^{3}=28, \\
a_{1} q+a_{1} q^{3}=2\left(a_{1} q^{2}+2\right),
\end{array}
$$

From (1) and (2), eliminating the constant, we get $6 a_{1} q^{3}-15 a_{1} q^{2}+6 a_{1} q=0$, which simplifies to $2 q^{2}-5 q+2=0$. Solving this, we get $q=2$ or $q=\frac{1}{2}$. Substituting into (1), we get $a_{1}=2$ or $a_{1}=32$. Since the sequence $\left\{a_{n}\right\}$ is monotonically increasing, we have $a_{1}=2$, $q=2$. Therefore, the general term of the sequence $\left\{a_{n}\right\}$ is $a_{n}=2^{n}$.
(2) From (1), we have $b_{n}-a_{n} \log _{\frac{1}{2}} a_{n}=-n \cdot 2^{n}$, thus
$$
S_{n}=b_{1}+b_{2}+\cdots+b_{n}=-\left(1 \times 2+2 \times 2^{2}+3 \times 2^{3}+\cdots+n \cdot 2^{n}\right),
$$

Let $T_{n}=1 \times 2+2 \times 2^{2}+3 \times 2^{3}+\cdots+n \cdot 2^{n}$, then
$$
2 T_{n}=1 \times 2^{2}+2 \times 2^{3}+3 \times 2^{4}+\cdots+(n-1) \times 2^{n}+n \times 2^{n+1},
$$

Subtracting these two equations, we get $-T_{n}=1 \times 2+1 \times 2^{2}+\cdots+1 \times 2^{n}-n \cdot 2^{n+1}$
$$
=2^{n+1}-2-n \cdot 2^{n+1} \text {, so } S_{n}=-(n-1) 2^{n+1}-2 \text {. }
$$

To make $S_{n}+n \cdot 2^{n+1}>50$ hold, we need $-(n-1) \cdot 2^{n+1}-2+n \cdot 2^{n+1}>50,2^{n}>26$. Since $2^{4}=16<26<32=2^{5}$, and the function $y=2^{x}$ is monotonically increasing, the smallest positive integer $n$ that satisfies the condition is 5.