Let $$\lim _\limits{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right.$$ $$\left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)$$ be $$\frac{\pi}{k}$$, using only the principal values of the inverse trigonometric functions. Then $$\mathrm{k}^2$$ is equal to _________.

Let $$f:(0, \pi) \rightarrow \mathbf{R}$$ be a function given by $$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0< x<\frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{\mathrm{b}}{\mathrm{a}}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$$

where $$\mathrm{a}, \mathrm{b} \in \mathbf{Z}$$. If $$f$$ is continuous at $$x=\frac{\pi}{2}$$, then $$\mathrm{a}^2+\mathrm{b}^2$$ is equal to _________.

The remainder when $$428^{2024}$$ is divided by 21 is __________.

If a function $$f$$ satisfies $$f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$$ for all $$\mathrm{m}, \mathrm{n} \in \mathbf{N}$$ and $$f(1)=1$$, then the largest natural number $$\lambda$$ such that $$\sum_\limits{\mathrm{k}=1}^{2022} f(\lambda+\mathrm{k}) \leq(2022)^2$$ is equal to _________.