$(3)解：如圖所示，以AB為邊在AB上方作Rt△BAE，且∠EAB=90°，∠EBA=30°，連接BE，EA，ED，EC，$
$在Rt△BDC中，∠DBC=30°，$
$在Rt△BAE中，∠EAB=90°，∠EBA=30°，∴△ABE∽△CBD$
$∴∠ABE=∠CBE$
$∴∠ABC=∠EBD$
$\frac {DB}{EB}=\frac {BC}{BA}$
$∴△BDE∽△BCA$
$∴\frac {DE}{AC}=\frac {BD}{BC}=\frac {2\sqrt{3}}{3} $
$∵AC=2$
$∴DE=\frac {4\sqrt{3}}{3}$
$在Rt△AEB中，AB=4$
$AE=AB×tan∠EBA=4×\frac {\sqrt{3}}{3}=\frac {4\sqrt{3}}{3}$
$∴D在以E為圓心，\frac {4\sqrt{3}}{3}為半徑的圓上運(yùn)動(dòng)$
$∴當(dāng)點(diǎn)A，E，D三點(diǎn)共線時(shí)，AD的值最大，此時(shí)$
$如圖所示，則AD=AE+DE=\frac {8\sqrt{3}}{3}$
$在Rt△ABD中$
$BD=\sqrt{AB^2+AD^2}=\sqrt{4^2+(\frac {8\sqrt{3}}{3})^2}=\frac {4\sqrt{21}}{3}$
$∴cos ∠BDA=\frac {AD}{BD}=\frac {\frac {8\sqrt{3}}{3}}{\frac {4\sqrt{21}}{3}}=\frac {2\sqrt{7}}{7}$
$sin∠BDA=\frac {AB}{BD}=\frac {4}{\frac {4\sqrt{21}}{3}}=\frac {\sqrt{21}}{7}$
$∵∠BEA=90°$
$∴∠BED=90°$
$∵△ABC∽△EBD$
$∴∠BDE=∠BCA$
$過點(diǎn)A做AF⊥BC于點(diǎn)F$
$∴CF=AC×cos∠ACB=2×\frac {2\sqrt{7}}{7}=\frac {4\sqrt{7}}{7}$
$AF=AC×sin∠ACB=\frac {2\sqrt{21}}{7}$
$∵∠DBC=30°$
$∴BC=\frac {\sqrt{3}}{2}BD=\frac {\sqrt{3}}{2}×\frac {4\sqrt{11}}{3}=2\sqrt{7}$
$∴BF=BC-CF=2\sqrt{7}-\frac {4\sqrt{7}}{7}=\frac {10\sqrt{7}}{7}$
$在Rt△AFB中，tan ∠CBA=\frac {AF}{BF}=\frac {\frac {2\sqrt{21}}{7}}{\frac {10\sqrt{7}}{7}}=\frac {\sqrt{3}}{5}$
