解：
(1) 证明：连接$EF。$
$\because AE$平分$∠ BAC，$
$\therefore ∠ BAE = ∠ CAE。$
$\because FE = FA，$
$\therefore ∠ FEA = ∠ BAE。$
$\therefore ∠ CAE = ∠ FEA。$
$\therefore EF // AC。$
$\therefore ∠ FEB = ∠ C = 90°。$
$\therefore EF ⊥ BC。$
又$\because EF$是$\odot F$的半径，
$\therefore BC$是$\odot F$的切线。
(2) 连接$FD。$
$\because$ 点$A$的坐标为$(0,-1)，$点$D$的坐标为$(2,0)，$
$\therefore OA = 1，$$OD = 2。$
设$\odot F$的半径为$R，$则$OF = R - 1。$
在$\mathrm{Rt}△ FOD$中，由勾股定理，得$OF^2 + OD^2 = FD^2，$
$\therefore (R - 1)^2 + 2^2 = R^2，$
解得$R = 2.5。$
$\therefore \odot F$的半径为2.5。
(3) $AG = AD + 2CD，$证明如下：
过点$E$作$EM ⊥ AG，$垂足为$M。$
$\because ∠ C = 90°，$
$\therefore EC ⊥ AC。$
又$\because AE$平分$∠ BAC，$$EM ⊥ AG，$
$\therefore EM = EC。$
在$\mathrm{Rt}△ AEM$和$\mathrm{Rt}△ AEC$中，
$\begin{cases} AE = AE \\ EM = EC \end{cases}$
$\therefore \mathrm{Rt}△ AEM ≌ \mathrm{Rt}△ AEC。$
$\therefore AM = AC。$
$\therefore AG - MG = AD + CD。$
连接$GE,ED。$
$\because ∠ BAE = ∠ CAE，$
$\therefore \overset{\frown}{EG} = \overset{\frown}{ED}。$
$\therefore EG = ED。$
又$\because EM = EC，$
$\therefore \mathrm{Rt}△ GEM ≌ \mathrm{Rt}△ DEC。$
$\therefore MG = CD。$
$\therefore AG - CD = AD + CD，$即$AG = AD + 2CD。$
