If $\mathrm{f}(x)= \begin{cases}\frac{8^x-4^x-2^x+1^x}{x^2} & , \text { if } x>0 \\ \mathrm{e}^x \sin x+i x+\lambda \log 4 & , \text { if } x \leqslant 0, i \in \mathbb{R}\end{cases}$ is continuous at $x=0$, then the value of $500 e^\lambda$ is

Ensuring continuity at $x=0$ requires:

$\lim_{x\to 0^-} \mathrm{f}(x) = \lim_{x\to 0^+} \mathrm{f}(x) = \mathrm{f}(0)$

The left-hand limit evaluates to $\lambda \log 4$, as $\mathrm{e}^x \to 1$, $\sin x \to 0$, and $i x \to 0$.

The right-hand limit transforms to $\lim_{x\to 0^+} \frac{(4^x-1)(2^x-1)}{x^2}$. Applying the standard limit $\lim_{x\to 0} \frac{a^x-1}{x} = \log a$ yields $(\log 4)(\log 2)$.

Since $\log 4 = 2\log 2$, this simplifies to $2(\log 2)^2$.

Equating both sides: $\lambda \log 4 = 2(\log 2)^2$. Substituting $\log 4 = 2\log 2$ gives $\lambda (2\log 2) = 2(\log 2)^2$, so $\lambda = \log 2$.

The expression $500\mathrm{e}^{\lambda}$ becomes $500\mathrm{e}^{\log 2} = 500 \cdot 2 = 1000$.

$\boxed{1000}$