Since, $\alpha$ and $\beta$ are roots of the quadratic equation
$$
x^2-4 x+5=0
$$


Now, $\left(\alpha^2+\beta\right)+\left(\alpha+\beta^2\right)=\left(\alpha^2+\beta^2\right)+(\alpha+\beta)$
$$
\begin{aligned}
& =(\alpha+\beta)^2-2 \alpha \beta+(\alpha+\beta) \\
& =16-10+4=10
\end{aligned}
$$
and $\left(\alpha^2+\beta\right)$
$$
\begin{aligned}
& \left(\alpha+\beta^2\right)=\alpha^3+\alpha^2 \beta^2+\beta \alpha+\beta^3 \\
& =\alpha^3+\beta^3+\alpha \beta(\alpha \beta+1) \\
& =(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)+\alpha \beta(\alpha \beta+1) \\
& =(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]+\alpha \beta(\alpha \beta+1) \\
& =4[16-15]+5(5+1) \\
& =4+30=34
\end{aligned}
$$
So, the quadratic equation whose roots are $\left(\alpha^2+\beta\right)$ and $\left(\alpha+\beta^2\right)$ is
$$
\begin{aligned}
x^2-\left(\alpha^2+\beta+\alpha+\beta^2\right) x+\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right) & =0 \\
\Rightarrow \quad x^2-10 x+34 & =0
\end{aligned}
$$