1).......
\bg_white \100dpi Assume\: vertex\: C\: to\: be \:origin\:\vec{0},\:and \:vertices \:B\:, A\:, D\:, as\: \vec{B},\:\vec{A},\:\vec{D}\\So\:\vec{A}-\vec{D}=\vec{a}\\\vec{B}-\vec{A}=\vec{b} \\\vec{X}=\frac{\vec{D}+\vec{B}}{2}\\\\\vec{Y}=\frac{\vec{A}}{2}\\and\:we\:know\:\vec{B}-\vec{C}=k\vec{a}\\its\:given\left | \vec{a} \right |=17\:and\:\left | \vec{XY} \right | =4 \\we\: get\\\left | \frac{\vec{A}-\vec{D}-\vec{B}}{2} \right |=4\ \\\left | \vec{A}-\vec{D} \right |=17\\
\bg_white \100dpi Now\: sqauring\:we\:get\\ \left | {\vec{A}-\vec{D}} \right |^{2}+\left | \vec{B} \right |^{2}-2(\vec{A}-\vec{D}).\vec{B}=64\\ we\:know\:\vec{B}=k.\vec{a} \\and\ \vec{A}-\vec{D}=\vec{a}\\ so\: \left | \vec{a} \right|^2+k^2.\left | \vec{a} \right|^2-2k\vec{a}.\vec{a}=64\\ 17^2+k^2.17^2-2.k.17^2=64\\ 17^2(1+k^2-2k)=64\\ (k-1)^2=(\frac{8}{17})^2\\ \therefore k=9/17\: or\: 25/17