Five. (Full marks 13 points) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-7 x+8=0$, and $\alpha>\beta$. Without solving the equation, use the relationship between roots and coefficients to find the value of $\frac{2}{\alpha}+3 \beta^{2}$.

---

The above text translated into English, preserving the original text's line breaks and format, is as follows:

Five. (Full marks 13 points) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-7 x+8=0$, and $\alpha>\beta$. Without solving the equation, use the relationship between roots and coefficients to find the value of $\frac{2}{\alpha}+3 \beta^{2}$.
Given the relationships between the roots and coefficients $\alpha+\beta=7, \alpha \beta=8$, we have
$$
\begin{array}{l}
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=7^{2}-2 \times 8=33, \\
(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=17 .
\end{array}
$$

Since $\alpha>\beta$, we have $\alpha-\beta=\sqrt{17}, \beta-\alpha=-\sqrt{17}$.
Let $A=\frac{2}{\alpha}+3 \beta^{2}$, and $B=\frac{2}{\beta}+3 \alpha^{2}$, then
$$
\begin{aligned}
A+B & =2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3\left(\alpha^{2}+\beta^{2}\right) \\
& =\frac{2(\alpha+\beta)}{\alpha \beta}+3\left(\alpha^{2}+\beta^{2}\right) \\
& =\frac{2 \times 7}{8}+3 \times 33=100 \frac{3}{4} . \\
A-B & =2\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)+3\left(\beta^{2}-\alpha^{2}\right) \\
& =\frac{2(\beta-\alpha)}{\alpha \beta}+3(\beta+\alpha)(\beta-\alpha) \\
& =\frac{-2 \sqrt{17}}{8}+3 \times 7 \times(-\sqrt{17}) \\
& =-\frac{85}{4} \sqrt{17} .
\end{aligned}
$$

Thus, $A=\frac{1}{2}\left(100 \frac{3}{4}-\frac{85}{4} \sqrt{17}\right)$
$$
=\frac{1}{8}(403-85 \sqrt{17})
$$