解：
(1) $AB// CF，$理由如下：
$\because DE// BC，$
$\therefore ∠ ADE = ∠ ABC。$
又$\because ∠ ADE + ∠ BCF = 180°，$
$\therefore ∠ ABC + ∠ BCF = 180°，$
$\therefore AB// CF$（同旁内角互补，两直线平行）。
(2) 过点$E$作$EM// AB，$
$\because AB// CF，$
$\therefore EM// AB// CF，$
$\therefore ∠ BEM = ∠ ABE = 40°，$$∠ MEC = ∠ ACF = 60°，$
$\therefore ∠ BEC = ∠ BEM + ∠ MEC = 40° + 60° = 100°。$
(3) 补图：延长$FC$至点$G，$连接$BG。$
$\because BE$平分$∠ ABG，$$∠ ABE=40°，$
$\therefore ∠ ABG = 2∠ ABE = 80°，$即$∠ ABC + ∠ GBC = 80°。$
$\because AB// CF，$
$\therefore ∠ ABC + ∠ BCF = 180°，$
又$\because ∠ BCF = ∠ ECB + ∠ ACF = ∠ ECB + 60°，$
$\therefore ∠ ABC + ∠ ECB = 120°。$
设$∠ GBC=2x，$则$∠ ECB=7x，$代入$∠ ABC + ∠ GBC = 80°$得$∠ ABC=80°-2x，$将其代入$∠ ABC + ∠ ECB = 120°$：
$80° - 2x + 7x = 120°$
解得$x=8°，$
$\therefore ∠ ABC = 80° - 2×8° = 64°。$
$\because DE// BC，$$AB// CF，$
$\therefore$ 四边形$DBCF$是平行四边形，
$\therefore ∠ F = ∠ ABC = 64°。$
答：$∠ F$的度数为$64°。$
