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证明：
(1) 设$AB$交$CD$于点$H，$连接$OC，$$OD。$
$\because AB$是$\odot O$的直径，$AB\perp CD，$$\therefore CH=DH$（垂径定理）。
在$\text{Rt}\triangle COH$和$\text{Rt}\triangle DOH$中，$\begin{cases}OC=OD\\OH=OH\end{cases}，$$\therefore \text{Rt}\triangle COH\cong\text{Rt}\triangle DOH$（HL），$\therefore \angle COH=\angle DOH，$$\therefore \overset{\frown}{BC}=\overset{\frown}{BD}，$$\therefore \angle BAC=\angle BAD。$
$\because \angle BOD$是$\overset{\frown}{BD}$所对的圆心角，$\angle BAC$是$\overset{\frown}{BC}$所对的圆周角，且$\overset{\frown}{BC}=\overset{\frown}{BD}，$$\therefore \angle BOD=2\angle BAC，$即$\angle BOD=2\angle A。$
(2) 连接$OC，$$AD。$
$\because F$为$AC$的中点，$OA=OC，$$\therefore OF\perp AC$（等腰三角形三线合一），$\therefore \angle AFO=90^\circ。$
$\because AB\perp CD，$$\therefore \angle AHD=90^\circ，$$\therefore \angle AFO=\angle AHD，$$\therefore OF// CD。$
$\because OA=OB，$$F$为$AC$中点，$\therefore OF$是$\triangle ABC$的中位线，$\therefore OF=\frac{1}{2}BC。$
设$\odot O$半径为$r，$则$OD=r。$由
(1)设$\angle A=\alpha，$则$\angle BOD=2\alpha，$$\angle AOD=180^\circ-2\alpha。$
$\because OA=OD，$$\therefore \angle OAD=\frac{180^\circ-(180^\circ-2\alpha)}{2}=\alpha，$$\therefore \angle OAD=\angle A，$$\therefore AD=OD=r。$
$\because AB$是直径，$\therefore \angle ADB=90^\circ，$$\therefore \angle ABD=90^\circ-\alpha。$
$\because \overset{\frown}{BC}=\overset{\frown}{BD}，$$\therefore BC=BD。$
$\because CE\perp DB，$$\therefore \angle E=90^\circ=\angle ADB，$$\therefore CE// AD。$
又$\because OF// CD，$可证$OC// DB，$$\therefore \angle OCE=\angle E=90^\circ，$即$OC\perp CE。$
$\because OC$是$\odot O$半径，$\therefore$直线$CE$为$\odot O$的切线。
证明：
(1) $\because AB=AD，$$\therefore \overset{\frown}{AB}=\overset{\frown}{AD}，$$\therefore \angle ACB=\angle ACD。$
$\because$ 四边形$ABCD$内接于圆，$\therefore \angle DAB+\angle DCB=180^\circ。$
$\because \angle BAD=2\angle DFC，$$\therefore 2\angle DFC+\angle DCB=180^\circ，$即$\angle DFC+\frac{1}{2}\angle DCB=90^\circ。$
$\because \angle ACB=\angle ACD=\frac{1}{2}\angle DCB，$$\therefore \angle DFC+\angle ACD=90^\circ，$$\therefore \angle CDF=90^\circ，$即$CD\perp DF。$
(2) 过点$F$作$FG\perp BC$于$G，$则$\angle FGC=90^\circ。$
$\because \angle FDC=90^\circ，$$\therefore \angle FGC=\angle FDC。$
又$\because \angle FCG=\angle FCD，$$FC=FC，$$\therefore \triangle FGC\cong\triangle FDC$（AAS），$\therefore CD=CG，$$\angle CFG=\angle CFD。$
$\because \angle BFC=2\angle DFC，$$\therefore \angle BFC=2\angle CFG，$$\therefore \angle BFG=\angle CFG。$
又$\because FG=FG，$$\angle FGB=\angle FGC=90^\circ，$$\therefore \triangle BFG\cong\triangle CFG$（ASA），$\therefore BG=CG，$$\therefore BC=BG+CG=2CG=2CD。$
解:(1) 设圆锥母线长为$l，$底面直径为$d，$由题意剖面是等边三角形，$\therefore l=d，$底面半径$r=\frac{d}{2}=\frac{l}{2}。$
圆锥侧面积$S_{\text{侧}}=\pi rl=\pi\cdot\frac{l}{2}\cdot l=\frac{1}{2}\pi l^2。$
侧面展开图弧长$l_{\text{弧}}=\pi d=\pi l，$设圆心角为$\alpha，$则$\alpha=\frac{l_{\text{弧}}}{l}\cdot\frac{180^\circ}{\pi}=\frac{\pi l}{l}\cdot\frac{180^\circ}{\pi}=180^\circ。$
由剖面底边长$30\,\text{cm}，$得$d=30\,\text{cm}，$$l=30\,\text{cm}，$$\therefore S_{\text{侧}}=\frac{1}{2}\pi\times30^2=450\pi\,\text{cm}^2，$圆心角$\alpha=180^\circ。$
(2) 裁剪方法：直径$60\,\text{cm}$的圆形铁皮，上半圆（圆心角$180^\circ，$半径$30\,\text{cm}$）作为圆锥侧面展开图，下半圆中剪一个半径$7.5\,\text{cm}$的小圆作为底面盖。（图形示意：上半圆和下半圆内的小圆）
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