8. Given $a_{1}=1, a_{n+1}=\lambda a_{n}^{2}+2\left(n \in \mathbf{N}^{*}\right)$, if the sequence $\left\{a_{n}\right\}$ is bounded above, i.e., there exists a constant $M$, such that $a_{n} \leqslant M$ for all $n \in \mathbf{N}^{*}$, then the maximum value of the real number $\lambda$ is $\qquad$.
Consider $\lambda>0$ only. Since $a_{n+1}-a_{n}=\lambda a_{n}^{2}-a_{n}+2=\lambda\left(a_{n}-\frac{1}{2 \lambda}\right)^{2}+2-\frac{1}{4 \lambda}$, when $\lambda>\frac{1}{8}$, the sequence $\left\{a_{n}\right\}$ is monotonically increasing and has no upper bound; when $0<\lambda \leqslant \frac{1}{8}$, we will prove that for $n \in \mathbf{N}^{*}$, $0<a_{n} \leqslant 4$. For $n=1$, the conclusion holds; assume that for $n=k$, the conclusion holds, i.e., $0<a_{k} \leqslant 4$. For $n=k+1$, $0<a_{k+1}=\lambda a_{n}^{2}+2 \leqslant \frac{1}{8} \cdot 4^{2}+2=4$, then for $n=k+1$, the conclusion also holds.