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解：??$(1)$??設(shè)正方形??$A、$????$B$??的邊長(zhǎng)分別為??$a、$????$b，$??
由題意，得??$\begin{cases}{ 2a=3b }\\{ a+b=10 }\end{cases}$??
解得??$\begin{cases}{ a=6 }\\{ b=4}\end{cases}$??
∴正方形??$A、$????$B$??的邊長(zhǎng)分別為??$6、$????$4.$??
??$(2)$??設(shè)正方形??$C、$????$D$??的邊長(zhǎng)分別為??$c、$????$d，$??則??$(c-d)^2=4，$??即??$c^2-2cd+d^2=4$??
由圖③，得??$(c+d)^2-c^2-d^2=48，$??即??$2dc=48$??
∴??$c^2+d^2-48=4，$??
∴??$c^2+d^2=52，$??即正方形??$C、$????$D$??的面積和為??$52.$??
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解：???$(1)②$???存在，過(guò)點(diǎn)???$D$???作???$DG⊥AC$???于點(diǎn)???$G，$
???作???$DH⊥BC$???于點(diǎn)???$H$???
∵點(diǎn)???$D$???到???$AC、$??????$BC$???的距離分別為???$4、$??????$3$???
∴???$DG=4，$??????$DH=3$???
由題意得，???$BP=2t，$??????$CQ=t$???
∵???$BC=8$???
∴???$CP=8-2t$???
∴???$S_{△CDP}=\frac 1 2CP·DH=\frac 1 2(8-2t)×3=12-3t，$???
???$S_{△CDQ}=\frac 1 2CQ·DG=\frac 1 2t×4=2t$???
令???$12-3t=2t，$???得
???$t=2.4$???
∴當(dāng)???$t=2.4$???時(shí)，???$△CDP$???與???$△CDQ$???面積相等
???$(2)$???結(jié)論：???$∠CFB=2∠ACD+∠ABE$???
證明：過(guò)點(diǎn)???$D$???作???$DM⊥BC$???于點(diǎn)???$M，$???
∵???$DM⊥BC$???
∴???$∠DMC=∠DMB=90°$???
∵???$∠DBC=∠DCB$???
∴???$∠CDM=∠BDM，$???即???$∠CDB=2∠CDM$???
∵???$∠ACB=∠DMB=90°$???
∴???$AC//DM$???
∴???$∠ACD=∠CDM$???
∴???$∠CDB=2∠ACD$???
∴???$∠CFB=∠CDB+∠ABE=2∠ACD+∠ABE$???
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