Use Pythagoras on the three relevant faces: \(AH=\sqrt{6^2+6^2}=\sqrt{72},\quad AF=\sqrt{8^2+6^2}=10,\quad HF=\sqrt{8^2+6^2}=10\) cm. Cosine rule in triangle AHF gives \(\cos\angle HAF=\frac{72+100-100}{2\sqrt{72}(10)}\). Therefore \(\angle HAF=64.9^\circ\).