Sure, here is the translated text:

```
II. (40 points) Find all positive integer quadruples $(p, q, x, y)$, where $p<q$, and $(p, q)=1$, such that
$$
3^{x \cos \frac{8}{q} \pi}+1=2^{y} \text {. }
$$
```
From the given, we have $3^{x \cos \frac{2}{q} \pi}=2^{y}-1$.
Thus, $x \cos \frac{p_{-\pi \in}}{q}$.
(1) If $x \cos \frac{p}{q} \pi=0$, then $y=1$.
(2) If $x \cos \frac{p}{q} \pi \neq 0$, then $x \cos \frac{p}{q} \pi \in \mathbf{N}_{+}$.

Therefore, $3^{x \cos \frac{p}{q^{\prime} \pi}}+1=2^{y} \equiv 1(\bmod 3)$
$\Rightarrow 2^{y} \equiv(-1)^{y} \equiv 1(\bmod 3)$
$\Rightarrow y$ is even.
Let $y=2 y_{0}$. Then
$3^{x \cos \frac{2}{q^{\pi}}}=2^{2 y_{0}}-1=\left(2^{y_{0}}-1\right)\left(2^{y_{0}}+1\right)$.
Thus, it can only be $\left\{\begin{array}{l}2^{y_{0}}-1=3^{m}, \\ 2^{y_{0}}+1=3^{n}\end{array},(m, n \in \mathbf{N})\right.$.
Subtracting the two equations gives $3^{n}-3^{m}=2$.
Therefore, it can only be $n=1, m=0$, i.e., $y_{0}=1$.
Hence, $3^{x \cos } \frac{p}{!}=3$.
Clearly, $\cos \frac{p}{q} \pi$ should be a rational number.
Let $\theta=\frac{p}{q} \pi$. Then $\cos q \theta= \pm 1$.
Next, we prove:
$2 \cos q \theta=f(2 \cos \theta)$,
where $f(x)$ is a polynomial with integer coefficients and leading coefficient 1.
When $q=1$ or 2, it is obviously true.
Assume that (1) holds for $q \leqslant n-1$.
Then when $q=n$, we have
$\cos n \theta=2 \cos (n-1) \theta \cdot \cos \theta-\cos (n-2) \theta$.
Therefore, by mathematical induction, the proposition holds.
Also, $f(2 \cos \theta)=2 \cos q \theta$, whose constant term is $\pm 2$ or 0.
Let the rational root of this integer coefficient polynomial be $\frac{s}{t}$. Then
$$
\left(\frac{s}{t}\right)^{q}+a_{1}\left(\frac{s}{t}\right)^{q-1}+\cdots \pm 2=0,
$$

Thus, $s|2, t| 1$.
(i) If $q=1$, then $\cos \frac{p}{q} \pi= \pm 1$.
(ii) If $q>1$, it can only be $s= \pm 1, t= \pm 1$.
In this case, $2 \cos \frac{p}{q} \pi= \pm 1$.
Hence, $\cos \frac{p}{q} \pi= \pm \frac{1}{2}$.
(iii) If the constant term of the polynomial is 0, then $\cos \frac{p}{q} \pi$ can

be 0. Otherwise, after canceling $2 \cos \frac{p}{q} \pi$, the leading term is 1, and similarly, it can be proven.
Therefore, the solutions are
$$
(p, q, 0,1),(1,2, x, 1),(1,3,2,2) \text {, }
$$

where $p, q, x \in \mathbf{N}_{+}$.