$解：?(2)?存在，過?B'?作?MN//AB，?交?AD，??BC?于點?M，??N，?$
$過?E?作?EH//AD，?交?MN?于?H，?$
$∵?AD//BC，??MN//AB，?$
$∴四邊形?ABNM?是平行四邊形，$
$又∵?∠A＝90°，?$
$∴四邊形?ABNM?是矩形，$
$同理可得：四邊形?AEHM?是矩形．$
$①如圖?1，?若點?B'?在?AD?下方，則?B'M＝3\ \mathrm {cm}，??B'N＝3\ \mathrm {cm}，?$
$∵?MH＝AE＝1(\mathrm {cm})，?$
$∴?B'H＝2(\mathrm {cm})，?$
$由折疊可得，?EB'＝EB＝5(\mathrm {cm})，?$
$∴?Rt△EB'H?中，?EH＝\sqrt {52-22}＝\sqrt {21}(\mathrm {cm})，?$
$∴?BN＝AM＝EH＝\sqrt {21}(\mathrm {cm})，?$
$∵?BP＝t，?$
$∴?PB'＝t，??PN＝\sqrt {21}-t，?$
$∵?Rt△PB'N?中，?B'P2＝PN2+B'N2，?$
$∴?t2＝(\sqrt {21}-t)2+32，?$
$解得?t＝\frac {5\sqrt {21}}{7}．?$
$②如圖?2，?若點?B'?在?AD?上方，則?B'M＝3\ \mathrm {cm}，??B'N＝9\ \mathrm {cm}，?$
$同理可得，?EH＝3\ \mathrm {cm}，?$
$∵?BP＝t，?∴?B'P＝t，??PN＝t﹣3，?$
$∵?Rt△PB'N?中，?B'P2＝PN2+B'N2，?$
$∴?t2＝(t-3)2+92，?$
$解得?t＝15．?$
$綜上所述，?t?的值為?\frac {5\sqrt {21}}{7}?秒或?15?秒．$
