8. If the inequality $x^{2}+1 \geqslant a x+b \geqslant \frac{3}{2} x^{\frac{2}{3}}$ holds for all $x$ in $[0,+\infty$ ) (where $a$ and $b$ are real numbers), then the range of values for $b$ is $\qquad$, and $a$ and $b$ satisfy $\qquad$.
8. $\left\{\begin{array}{l}\frac{2-\sqrt{2}}{4} \leqslant b \leqslant \frac{2+\sqrt{2}}{4} \\ \frac{1}{\sqrt{2 b}} \leqslant a \leqslant 2 \sqrt{1-b}\end{array}\right.$ Detailed explanation; Taking $x=0$, we get $1 \geqslant b \geqslant 0$; For $x>0$, we always have $a x+b \geqslant \frac{3}{2} x^{\frac{2}{3}}$, thus it must be that $a>0$. Let $f(x)=x^{2}-a x+1-b, x \in[0,+\infty)$, then $f^{\prime}(x)=2 x-a$, setting $f^{\prime}(x)=0$, we get $x=\frac{a}{2}$.
\begin{tabular}{|c|c|c|c|}
\hline$x$ & $\left(0, \frac{a}{2}\right)$ & $\frac{a}{2}$ & $\left(\frac{a}{2},+\infty\right)$ \\
\hline$f^{\prime}(x)$ & - & 0 & + \\
\hline$f(x) \because$ & $\downarrow$ & Minimum value & $\uparrow$ \\
\hline
\end{tabular}

Thus, $f\left(\frac{a}{2}\right) \geqslant 0 \Leftrightarrow-\frac{a^{2}}{4}+1-b \geqslant 0 \Leftrightarrow b \leqslant 1-\frac{a^{2}}{4}$. Let $g(x)=a x+b-\frac{3}{2} x^{\frac{2}{3}}, x \in[0,+\infty)$, then $g^{\prime}(x)=a-x^{-\frac{1}{3}}$, setting $g^{\prime}(x)=0$, we get $x=\frac{1}{a^{3}}$.
\begin{tabular}{|c|c|c|c|}
\hline$x$ & $\left(0, \frac{1}{a^{3}}\right)$ & $\frac{1}{a^{3}}$ & $\left(\frac{1}{a^{3}},+\infty\right)$ \\
\hline$g^{\prime}(x)$ & - & 0 & + \\
\hline$g(x)$ & $\downarrow$ & Minimum value & $\uparrow$ \\
\hline
\end{tabular}

Thus, $g\left(\frac{1}{a^{3}}\right) \geqslant 0 \Leftrightarrow b-\frac{1}{2 a^{2}} \geqslant 0 \Leftrightarrow b \geqslant \frac{1}{2 a^{2}}$. In summary, the conditions that the real numbers $a$ and $b$ must satisfy are
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ 0 \leqslant b \leqslant 1 , a > 0 } \\
{ \frac { 1 } { 2 a ^ { 2 } } \leqslant b \leqslant 1 - \frac { a ^ { 2 } } { 4 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 0 \leqslant b \leqslant 1 , a > 0 } \\
{ \frac { 1 } { 2 b } \leqslant a ^ { 2 } \leqslant 4 ( 1 - b ) }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 0 \leqslant b \leqslant 1 , } \\
{ \frac { 1 } { \sqrt { 2 b } } \leqslant a \leqslant 2 \sqrt { 1 - b } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 0 \leqslant b \leqslant 1 , } \\
{ \frac { 1 } { 2 b } \leqslant 4 ( 1 - b ) } \\
{ \frac { 1 } { \sqrt { 2 b } } \leqslant a \leqslant 2 \sqrt { 1 - b } } \\
{ b ^ { 2 } - b + \frac { 1 } { 8 } \leqslant 0 } \\
{ \frac { 1 } { \sqrt { 2 b } } \leqslant a \leqslant 2 \sqrt { 1 - b } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{2-\sqrt{2}}{4} \leqslant b \leqslant \frac{2+\sqrt{2}}{4} \\
\frac{1}{\sqrt{2 b}} \leqslant a \leqslant 2 \sqrt{1-b}
\end{array}\right.\right.\right.\right.\right.
\end{array}
$$

Alternative solution: As shown in the figure, the line $y=a x+b$ lies between the graphs of the functions $y=x^{2}+1, x \in [0,+\infty)$ and $y=\frac{3}{2} x^{\frac{2}{3}}, x \in [0,+\infty)$. Therefore, the necessary conditions are: $\left\{\begin{array}{l}a>0 \\ 0 \leqslant b \leqslant 1\end{array}\right.$; Next, we continue to explore the sufficient conditions.

Draw the tangent line $l_{1}: y=2 x_{0}\left(x-x_{0}\right)+x_{0}^{2}+1$ to the graph of the function $y=x^{2}+1, x \in [0,+\infty)$ at the point $(0, b)$, where $T_{1}\left(x_{0}, x_{0}^{2}+1\right)$ is the point of tangency. Substituting the coordinates of the point $(0, b)$, we get $b=1-x_{0}^{2}$, thus the slope of the tangent line $k_{1}=2 x_{0}=2 \sqrt{1-b}$.
Draw the tangent line $l_{2}: y=x_{0}^{-\frac{1}{3}}\left(x-x_{0}\right)+\frac{3}{2} x_{0}^{\frac{2}{3}}$ to the graph of the function $y=\frac{3}{2} x^{\frac{2}{3}}, x \in [0,+\infty)$ at the point $(0, b)$, where $T_{2}\left(x_{0}, \frac{3}{2} x_{0}^{\frac{2}{3}}\right)$ is the point of tangency. Substituting the coordinates of the point $(0, b)$, we get $b=\frac{1}{2} x_{0}^{\frac{2}{3}}$, thus the slope of the tangent line $k_{2}=x_{0}^{-\frac{1}{3}}=\frac{1}{\sqrt{2 b}}$. Therefore,
$$
\begin{array}{l}
\frac{1}{\sqrt{2 b}} \leqslant a \leqslant 2 \sqrt{1-b} . \\
\therefore \text { From }\left\{\begin{array} { l } 
{ a > 0 } \\
{ 0 \leqslant b \leqslant 1 } \\
{ \frac { 1 } { \sqrt { 2 b } } \leqslant a \leqslant 2 \sqrt { 1 - b } }
\end{array} \text { we derive the necessary and sufficient conditions as } \left\{\begin{array}{l}
\frac{2-\sqrt{2}}{4} \leqslant b \leqslant \frac{2+\sqrt{2}}{4} \\
\frac{1}{\sqrt{2 b}} \leqslant a \leqslant 2 \sqrt{1-b}
\end{array} .\right.\right.
\end{array}
$$