8.
\((1)\)已知\(a+b=3\), \(ab=2\), 则\( \dfrac{1}{2}{a}^{3}b+{a}^{2}{b}^{2}+ \dfrac{1}{2}a{b}^{3} \)=___________\(;\)
\((2)\)关于\(x\)的方程\( \dfrac{2x+m}{x-2}=3 \)的解是正数,则\(m\)的取值范围是 .
\((3)\) 若等腰\(\triangle \)\(ABC\)的一边长为\(1\),另两边长恰好是关于\(x\)的方程\({x}^{2}-(k+2)x+2k=0 \)的两个根,则\(\triangle \)\(ABC\)的周长是 .
\((4)\)如图\(.\)在平面直角坐标系\(xOy\)中,直线\(y\)\(= \sqrt{3} \)\(x\)经过点\(A\),作\(AB\)\(⊥\)\(x\)轴于点\(B\),将\(\triangle \)\(ABO\)绕点\(B\)逆时针旋转\(60^{\circ}\)得到\(\triangle \)\(CBD\)\(.\)若点\(B\)的坐标为\((2\), \(0)\),则点\(C\)的坐标为___________\(;\)
\((5)\)已知:如图,在正方形\(ABCD\)外取一点\(E\),连接\(AE\),\(BE\),\(DE\)\(.\)过点\(A\)作\(AE\)的垂线\(AP\)交\(DE\)于点\(P\)\(.\)若\(AE\)\(=\)\(AP\)\(=1\),\(PB\)\(=\)\( \sqrt{5} \) 下列结论:
\(①\triangle \)\(APD\)≌\(\triangle \)\(AEB\); \(②\)点\(B\)到直线\(AE\)的距离为\( \sqrt{2} \); \(③\)\(EB⊥ED \);
\(④\)\({{S}_{\triangle }}_{APD}+{{S}_{\triangle }}_{APE}=1+ \sqrt{6} \)\(.⑤\)\({S}_{正方形ABCD}=4+ \sqrt{6} \). 其中正确结论的序号是 .