答案：1. 2

答案：
2. $\sqrt{2}$ 点拨: 如答图, 连接 $AE,AC,CG,CF$. $\because$ 四边形 $DEFG$ 是正方形, $\therefore ∠ EDG = 90^{\circ},EF = DE = DG$. $\because$ 四边形 $ABCD$ 是正方形, $\therefore ∠ ADC = 90^{\circ},AD = CD$, $\therefore ∠ ADE = ∠ CDG$, $\therefore △ ADE ≌ △ CDG(SAS)$, $\therefore AE = CG$, $\therefore m_1 + m_2 + m_3 = EF + CF + AE$, $\therefore$ 当点 $A,E,F,C$ 在同一条直线上时, $EF + CF + AE$ 的值最小, 即 $m_1 + m_2 + m_3$ 的值最小, $\therefore m_1 + m_2 + m_3$ 的最小值为 $AC$ 的长. 在 $Rt△ ACD$ 中, $AC = \sqrt{2}AB = \sqrt{2}$, $\therefore m_1 + m_2 + m_3$ 的最小值为 $\sqrt{2}$.

答案：
3. 解: (1) 如答图①.
$\because$ 四边形 $ABCD$ 是正方形, $\therefore AB = BC, ∠ ABC = 90^{\circ}$.
$\because BE = BA, \therefore AB = BE = BC$.
设 $∠ BAE = ∠ BEA = x^{\circ}, ∠ BEC = ∠ BCE = y^{\circ}$.
$\because$ 四边形 $ABCE$ 的内角和为 $360^{\circ}$,
$\therefore 2x + 2y + 90 = 360$,
$\therefore x + y = 135, \therefore ∠ AEC = 135^{\circ}, \therefore ∠ CEF = 45^{\circ}$.

(2) $AF = \sqrt{2}DF + CF$, 证明如下:
如答图②, 作 $DH ⊥ DF$, 交 $AF$ 于点 $H$,
$\therefore ∠ ADH = ∠ CDF = 90^{\circ} - ∠ HDC$.
$\because ∠ EFC = 90^{\circ}, ∠ CEF = 45^{\circ}$,
$\therefore ∠ CEF = ∠ FCE = 45^{\circ}, \therefore △ EFC$ 是等腰直角三角形,
$\therefore EF = FC$.
$\because ∠ DAB = 90^{\circ}$, 设 $∠ BAE = ∠ BEA = m, ∠ BEC = ∠ BCE = n, \therefore ∠ DAH = 90^{\circ} - m$.
$\because ∠ DCE = 90^{\circ} - n, \therefore ∠ FCD = 45^{\circ} - (90^{\circ} - n) = n - 45^{\circ}$.
又 $\because m + n = 135^{\circ}, \therefore n = 135^{\circ} - m$,
$\therefore ∠ FCD = 90^{\circ} - m, \therefore ∠ DAH = ∠ DCF$.
在 $△ DAH$ 和 $△ DCF$ 中, $\{\begin{array}{l} ∠ DAH = ∠ DCF, \\ AD = CD, \\ ∠ ADH = ∠ CDF, \end{array} $
$\therefore △ DAH ≌ △ DCF(ASA)$,
$\therefore AH = CF, DH = DF$,
$\therefore △ DHF$ 是等腰直角三角形, $\therefore HF = \sqrt{2}DF$.
$\because AF = HF + AH, \therefore AF = \sqrt{2}DF + CF$.