$解：?(1)?由題意得?CM=t m，??AN=2tm?$
$∵?∠C=90° ?$
$∴?AB=\sqrt {BC^2+AC^2}=13m?$
$∵?∠AMN=∠ANM ?$
$∴?AM=AN?$
$∴當(dāng)?t?為?4s?時(shí)，?∠AMN=∠ANM?$
$?(2)?如圖，過(guò)點(diǎn)?N?作?NH⊥AC?于點(diǎn)?H?$
$∵?∠AHN=∠C=90°?$
$∴?NH//BC?$
$∴?△ANH∽△ABC?$
$∴?\frac {AN}{AB}=\frac {NH}{BC}，?即?\frac {2t}{13}=\frac {NH}5?$
$∴?NH=\frac {10t}{13}m?$
$∴?S_{△AMN}=\frac 12×AM×NH=-\frac 5{13}(t-6)^2+\frac {180}{13}?$
$當(dāng)?t?為?6s?時(shí)，?△AMN?的面積最大，最大為?\frac {180}{13}\mathrm {m^2}?$
