Let g(x)=(ln x)/(x) for x>0.

$g'(x)=\dfrac{1-\ln x}{x^2}$, positive on $(0,e)$ and negative on $(e,\infty)$, so $g$ increases up to $x=e$ and decreases afterwards; the maximum is $g(e)=\tfrac1e$. (A) true. (B) Since $\pi>e$, $g(\pi)<g(e)$, i.e. $\dfrac{\ln\pi}\pi<\dfrac1e$, so $e\ln\pi<\pi$ and $\pi^e<e^\pi$; the claim $\pi^e>e^\pi$ is false. (C) For $x>0$, $x^3=3^x\iff g(x)=g(3)=\dfrac{\ln3}3>0$. Since $g$ is strictly increasing on $(0,e)$ from $-\infty$ to $\tfrac1e$ and strictly decreasing on $(e,\infty)$, the value $g(3)<\tfrac1e$ is taken exactly once on each side: at $x=3$ and at some $x\approx2.48$. For $x\le0$, $x^3\le0<3^x$, so no roots there. Exactly two: true. (D) $g(2024)>g(2025)$ (both exceed $e$), so $2025\ln2024>2024\ln2025$, i.e. $2024^{2025}>2025^{2024}$: true. Correct: A, C, D.