$(1)$如图$1$，$\because$四边形$ABCD$是矩形，$AB=8$，$AD=6$，

$\therefore \angle B=90^{\circ}$，$BC=AD=6$，

$\because O$、$E$分别是$AC$、$AB$的中点，

$\therefore EO$∥$BC$，$EO=\frac{1}{2}BC=3$，

$\therefore \angle AEO=\angle B=90^{\circ}$，

$\therefore \angle OEB=180^{\circ}-\angle AEO=90^{\circ}$，

$\because \angle EOF=90^{\circ}$，

$\therefore$四边形$BEOF$是矩形，

$\therefore BF=EO=3$，

$\because \frac{AE}{BE}=\frac{AO}{CO}=1$，

$\therefore BE=AE=\frac{1}{2}AB=4$，

$\therefore EF=\sqrt{B{F}^{2}+B{E}^{2}}=\sqrt{{3}^{2}+{4}^{2}}=5$，

$\therefore EF$的长是$5$.

$(2)$证明：如图$2$，连接$OB$，作$OG\bot BC$于点$G$，$OH\bot AB$于点$H$，

$\because \angle ABC=90^{\circ}$，$OA=OC$，

$\therefore OB=OA=OC=\frac{1}{2}AC$，

$\therefore AH=BH=\frac{1}{2}AB$，$CG=BG=\frac{1}{2}BC$，

$\therefore OH=\frac{1}{2}BC=CG=BG$，$OG=\frac{1}{2}AB=AH=BH,OG$∥$AB$，

$\therefore \frac{OH}{OG}=\frac{\frac{1}{2}BC}{\frac{1}{2}AB}=\frac{BC}{AB}$，$\angle GOH=\angle OHA=90^{\circ}$，

$\therefore \angle EOH=\angle FOG=90^{\circ}-\angle FOH$，

$\because \angle OHE=\angle OGF=90^{\circ}$，

$\therefore \triangle OHE$∽$\triangle OGF$，

$\therefore \frac{EH}{FG}=\frac{OH}{OG}=\frac{BC}{AB}$，

$\therefore AB\cdot EH=BC\cdot FG$，

设$OH=CG=BG=a$，$OG=AH=BH=b$，

$\therefore OE^{2}=\left(AE+b\right)^{2}+a^{2}=AE^{2}+2b\cdot AE+b^{2}+a^{2}$，$OF^{2}=\left(CF-a\right)^{2}+b^{2}=CF^{2}-2a\cdot CF+a^{2}+b^{2}$，

$\therefore OE^{2}+OF^{2}=AE^{2}+CF^{2}+2b\cdot AE+2b^{2}-2a\cdot CF+2a^{2}=AE^{2}+CF^{2}+2b\left(AE+b\right)-2a\left(CF-a\right)$，

$\because AB=2b$，$BC=2a$，$AE+b=EH$，$CF-a=FG$，

$\therefore 2b\left(AE+b\right)-2a\left(CF-a\right)=AB\cdot EH-BC\cdot FG=0$，

$\therefore EF^{2}=OE^{2}+OF^{2}=AE^{2}+CF^{2}$，

$\therefore EF^{2}-CF^{2}=AE^{2}$.

$(3)$如图$3$，作$OG\bot BC$于点$G$，$OH\bot AB$于点$H$，

$\because \angle OHB=\angle HBG=\angle OGB=90^{\circ}$，

$\therefore$四边形$OGBH$是矩形，

$\therefore \angle GOH=90^{\circ}$，

$\therefore \angle HOE=\angle GOF=90^{\circ}-\angle EOG$，

$\therefore \triangle OHE$∽$\triangle OGF$，

$\therefore \frac{EH}{FG}=\frac{OH}{OG}=\frac{BC}{AB}$，

$\because \angle ABC=90^{\circ}$，$\angle BAC=30^{\circ}$，$BC=1$，

$\therefore AE=AC=2BC=2$，

$\therefore AB=\sqrt{A{C}^{2}-B{C}^{2}}=\sqrt{{2}^{2}-{1}^{2}}=\sqrt{3}$，

$\therefore \frac{EH}{FG}=\frac{BC}{AB}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$，$AH=BH=OG=\frac{1}{2}AB=\frac{\sqrt{3}}{2}$，

$\therefore EH=\frac{\sqrt{3}}{3}FG=2-\frac{\sqrt{3}}{2}$，

$\therefore FG=2\sqrt{3}-\frac{3}{2}$，

$\because OH=BG=CG=\frac{1}{2}BC=\frac{1}{2}$，

$\therefore OE^{2}=(\frac{1}{2})^{2}+(2-\frac{\sqrt{3}}{2})^{2}=5-2\sqrt{3}$，$OF^{2}=(\frac{\sqrt{3}}{2})^{2}+(2\sqrt{3}-\frac{3}{2})^{2}=15-6\sqrt{3}$，

$\therefore EF^{2}=OE^{2}+OF^{2}=5-2\sqrt{3}+15-6\sqrt{3}=20-8\sqrt{3}$，

$\therefore$线段$EF^{2}$的值是$20-8\sqrt{3}$.

$(1)$如图$1$，$\because$四边形$ABCD$是矩形，$AB=8$，$AD=6$，

$\therefore \angle B=90^{\circ}$，$BC=AD=6$，

$\because O$、$E$分别是$AC$、$AB$的中点，

$\therefore EO$∥$BC$，$EO=\frac{1}{2}BC=3$，

$\therefore \angle AEO=\angle B=90^{\circ}$，

$\therefore \angle OEB=180^{\circ}-\angle AEO=90^{\circ}$，

$\because \angle EOF=90^{\circ}$，

$\therefore$四边形$BEOF$是矩形，

$\therefore BF=EO=3$，

$\because \frac{AE}{BE}=\frac{AO}{CO}=1$，

$\therefore BE=AE=\frac{1}{2}AB=4$，

$\therefore EF=\sqrt{B{F}^{2}+B{E}^{2}}=\sqrt{{3}^{2}+{4}^{2}}=5$，

$\therefore EF$的长是$5$.

$(2)$证明：如图$2$，连接$OB$，作$OG\bot BC$于点$G$，$OH\bot AB$于点$H$，

$\because \angle ABC=90^{\circ}$，$OA=OC$，

$\therefore OB=OA=OC=\frac{1}{2}AC$，

$\therefore AH=BH=\frac{1}{2}AB$，$CG=BG=\frac{1}{2}BC$，

$\therefore OH=\frac{1}{2}BC=CG=BG$，$OG=\frac{1}{2}AB=AH=BH,OG$∥$AB$，

$\therefore \frac{OH}{OG}=\frac{\frac{1}{2}BC}{\frac{1}{2}AB}=\frac{BC}{AB}$，$\angle GOH=\angle OHA=90^{\circ}$，

$\therefore \angle EOH=\angle FOG=90^{\circ}-\angle FOH$，

$\because \angle OHE=\angle OGF=90^{\circ}$，

$\therefore \triangle OHE$∽$\triangle OGF$，

$\therefore \frac{EH}{FG}=\frac{OH}{OG}=\frac{BC}{AB}$，

$\therefore AB\cdot EH=BC\cdot FG$，

设$OH=CG=BG=a$，$OG=AH=BH=b$，

$\therefore OE^{2}=\left(AE+b\right)^{2}+a^{2}=AE^{2}+2b\cdot AE+b^{2}+a^{2}$，$OF^{2}=\left(CF-a\right)^{2}+b^{2}=CF^{2}-2a\cdot CF+a^{2}+b^{2}$，

$\therefore OE^{2}+OF^{2}=AE^{2}+CF^{2}+2b\cdot AE+2b^{2}-2a\cdot CF+2a^{2}=AE^{2}+CF^{2}+2b\left(AE+b\right)-2a\left(CF-a\right)$，

$\because AB=2b$，$BC=2a$，$AE+b=EH$，$CF-a=FG$，

$\therefore 2b\left(AE+b\right)-2a\left(CF-a\right)=AB\cdot EH-BC\cdot FG=0$，

$\therefore EF^{2}=OE^{2}+OF^{2}=AE^{2}+CF^{2}$，

$\therefore EF^{2}-CF^{2}=AE^{2}$.

$(3)$如图$3$，作$OG\bot BC$于点$G$，$OH\bot AB$于点$H$，

$\because \angle OHB=\angle HBG=\angle OGB=90^{\circ}$，

$\therefore$四边形$OGBH$是矩形，

$\therefore \angle GOH=90^{\circ}$，

$\therefore \angle HOE=\angle GOF=90^{\circ}-\angle EOG$，

$\therefore \triangle OHE$∽$\triangle OGF$，

$\therefore \frac{EH}{FG}=\frac{OH}{OG}=\frac{BC}{AB}$，

$\because \angle ABC=90^{\circ}$，$\angle BAC=30^{\circ}$，$BC=1$，

$\therefore AE=AC=2BC=2$，

$\therefore AB=\sqrt{A{C}^{2}-B{C}^{2}}=\sqrt{{2}^{2}-{1}^{2}}=\sqrt{3}$，

$\therefore \frac{EH}{FG}=\frac{BC}{AB}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$，$AH=BH=OG=\frac{1}{2}AB=\frac{\sqrt{3}}{2}$，

$\therefore EH=\frac{\sqrt{3}}{3}FG=2-\frac{\sqrt{3}}{2}$，

$\therefore FG=2\sqrt{3}-\frac{3}{2}$，

$\because OH=BG=CG=\frac{1}{2}BC=\frac{1}{2}$，

$\therefore OE^{2}=(\frac{1}{2})^{2}+(2-\frac{\sqrt{3}}{2})^{2}=5-2\sqrt{3}$，$OF^{2}=(\frac{\sqrt{3}}{2})^{2}+(2\sqrt{3}-\frac{3}{2})^{2}=15-6\sqrt{3}$，

$\therefore EF^{2}=OE^{2}+OF^{2}=5-2\sqrt{3}+15-6\sqrt{3}=20-8\sqrt{3}$，

$\therefore$线段$EF^{2}$的值是$20-8\sqrt{3}$.