已知函数$f(x)=A\sin(\omega x+\phi )(A\gt 0,\omega \gt 0,|\phi |\lt \dfrac{\pi }{2})$，且$f\left(x\right)$图象的相邻两条对称轴之间的距离为$\dfrac{\pi }{2}$，请从条件①、条件②、条件③中任意选择两个作为已知条件作答.
条件①：$f\left(x\right)$的最小值为$-2$；
条件②：$f\left(x\right)$的图象的一个对称中心为$(\dfrac{5\pi }{12},0)$；
条件③：$f\left(x\right)$的图象经过点$(\dfrac{5\pi }{6},-1)$.
$(1)$求$f\left(x\right)$的解析式；
$(2)$在锐角$\triangle ABC$中，内角$A$，$B$，$C$所对的边分别为$a$，$b$，$c$，$A=\dfrac{\pi }{3}$，$b=f\left(A\right)$，求$\triangle ABC$面积的取值范围.

$(1)$$\\ f(x)=2\\sin(2x+\\dfrac{\\pi }{6})$;

$(2)$（$\\dfrac{\\sqrt{3}}{8},\\dfrac{\\sqrt{3}}{2})$

$(1)\because f\left(x\right)$图象的相邻两条对称轴之间的距离为$\dfrac{\pi }{2}$，$\therefore \dfrac{T}{2}=\dfrac{\pi }{2}$，即$T=\pi $，
$\therefore \omega =\dfrac{2\pi }{T}=\dfrac{2\pi }{\pi }=2$，$\therefore f\left(x\right)=A\sin \left(2x+\varphi \right)$.
若选①②，则$A=2$，$2×\dfrac{5\pi }{12}+\phi =k\pi $，$k\in \bf Z$，即$\phi =k\pi -\dfrac{5\pi }{6}$，$k\in\bf Z$，
$\because |\phi |\lt \dfrac{\pi }{2}$，$\therefore \phi =\dfrac{\pi }{6}$，$\therefore f(x)=2\sin(2x+\dfrac{\pi }{6})$；
若选①③，则$A=2$，$2\sin(2×\dfrac{5\pi }{6}+\phi )=-1$，即$\sin(\dfrac{5\pi }{3}+\phi )=-\dfrac{1}{2}$，
$\because |\phi |\lt \dfrac{\pi }{2}$，$\therefore \dfrac{7\pi }{6}\lt \dfrac{5\pi }{3}+\phi \lt \dfrac{13\pi }{6}$，$\therefore \dfrac{5\pi }{3}+\phi =\dfrac{11\pi }{6}$，得$\phi =\dfrac{\pi }{6}$，
$\therefore f(x)=2\sin(2x+\dfrac{\pi }{6})$；
若选②③，$2×\dfrac{5\pi }{12}+\phi =k\pi $，$k\in \bf Z$，即$\phi =k\pi -\dfrac{5\pi }{6}$，$k\in\bf Z$，
$\because |\phi |\lt \dfrac{\pi }{2}$，$\therefore \phi =\dfrac{\pi }{6}$，此时$f(x)=A\sin(2x+\dfrac{\pi }{6})$，
$\because f(\dfrac{5\pi }{6})=-1$，$A\sin(2×\dfrac{5\pi }{6}+\dfrac{\pi }{6})=-1$，$Asin\dfrac{11\pi }{6}=-1$，$A=2$，
$\therefore f(x)=2\sin(2x+\dfrac{\pi }{6})$.
$(2)$由$(1)$知，$f(x)=2\sin(2x+\dfrac{\pi }{6})$，$\because A=\dfrac{\pi }{3}$，
$\therefore b=f(\dfrac{\pi }{3})=2\sin(\dfrac{2\pi }{3}+\dfrac{\pi }{6})=1$，
$\because \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$，$\therefore c=\dfrac{b\sin C}{\sin B}=\dfrac{\sin(A+B)}{\sin B}=\dfrac{\sqrt{3}}{2\tan B}+\dfrac{1}{2}$，
$\therefore S=\dfrac{1}{2}bc\sin A=\dfrac{\sqrt{3}}{4}c=\dfrac{3}{8\tan B}+\dfrac{\sqrt{3}}{8}$，
$\because A+B \gt \dfrac{\pi }{2}$，$0 \lt B \lt \dfrac{\pi }{2}$，$\therefore \dfrac{\pi }{6} \lt B \lt \dfrac{\pi }{2}$，$\therefore \tan B \gt \dfrac{\sqrt{3}}{3}$，
$\therefore \dfrac{\sqrt{3}}{8}\lt {S}_{△ABC}\lt \dfrac{\sqrt{3}}{2}$，即$\triangle ABC$面积的取值范围是（$\dfrac{\sqrt{3}}{8},\dfrac{\sqrt{3}}{2})$.