Xét ΔBAC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(\dfrac{\left(4\sqrt{2}\right)^2+7^2-AC^2}{2\cdot4\sqrt{2}\cdot7}=cos45=\dfrac{\sqrt{2}}{2}\)
=>\(32+7^2-AC^2=\dfrac{\sqrt{2}}{2}\cdot2\cdot4\sqrt{2}\cdot7=28\)
=>\(AC=\sqrt{32+49-28}=\sqrt{32+21}=\sqrt{53}\left(cm\right)\)

