$(2)解：①如圖1，過點E作y軸平行線分別交AB、BD$
$于G、H$
$∵y=\frac {\sqrt{2}}{2}(x^2-3x-2)$
$∴A(0，-\sqrt{2})$
$∴AD=2\sqrt{2}，BD=4$
$∴AB=2\sqrt{6}$
$∴cos ∠ABD=\frac {\sqrt{6}}{3}$
$∴cos ∠FEG=\frac {\sqrt{6}}{3}$
$∴\frac {EF}{EG}=\frac {\sqrt{6}}{3}$
$∴EF=\frac {\sqrt{3}}{3}EG$
$AB直線解析式為y=\frac {\sqrt{2}}{2}x-\sqrt{2}$
$設E(m，\frac {\sqrt{2}}{2}m^2-\frac {3\sqrt{2}}{2}m-\sqrt{2})$
$∴G(m，\frac {\sqrt{2}}{2}m-\sqrt{2})$
$∴EG=-\frac {\sqrt{2}}{2}m^2+2\sqrt{2}m=-\frac {\sqrt{2}}{2}(m-2)^2+2\sqrt{2}$
$∴當m=2時，EG取得最大值2\sqrt{2}$
$∴EF的最大值為\frac {\sqrt{6}}{3}×2\sqrt{2}=\frac {4\sqrt{3}}{3}$
$②如圖2，已知tan∠ABC=\frac {\sqrt{2}}{2}，令$
$AC=\sqrt{2}，AB=2，在BC上截取AD=BD$
$∴∠ADC=2∠ABC$
$設CD=x，則AD=BD=2-x$
$則x^2+(\sqrt{2})^2=(2-x)^2$
$解得x=\frac12$
$∴tan ∠ADC=2\sqrt{2}$
$即tan(2∠ABC)=2\sqrt{2}$
$如圖3構造△AMF∽△FNE，相似比為AF：EF$
$又因為tan∠MFA=tan∠CBA=tan ∠FEN=\frac {\sqrt{2}}{2}$
$所以設AM=\sqrt{2}a，MF=2a$
$當∠FAE=2∠ABC時，tan ∠FAE=2\sqrt{2}$
$∴相似比為1:2\sqrt{2}$
$∴FN=2\sqrt{2}AM=4a$
$NE=2\sqrt{2}MF=4\sqrt{2}a$
$∴E(6a，-\sqrt{2}-3\sqrt{2}a)$
$代入拋物線得$
$a_1=\frac13，a_2=0(舍)$
$∴E點橫坐標為6a=2$
$當∠FEA=2∠ABC時，tan ∠FAE=2\sqrt{2}$
$相似比為2\sqrt{2}：1$
$∴FN=\frac {AM}{2\sqrt{2}}=\frac12a$
$NE=\frac {MF}{2\sqrt{2}}=\frac {\sqrt{2}}{2}a$
$∴E(\frac {5}{2}a，-\sqrt{2}+\frac {\sqrt{2}}{2}a)$
$代入拋物線求得a_1=\frac {34}{25}，a_2=0(舍)$
$∴E點橫坐標為\frac {5}{2}a=\frac {17}{5}$
$綜上所述，點E的橫坐標為2或\frac {17}{5}$
