Answer
$\theta=\pi$
Work Step by Step
$\theta=cos^{-1}\frac{u\cdot v}{|u||v|}=cos^{-1}\frac{(cos(\frac{\pi}{4})i+sin(\frac{\pi}{4})j)(cos(\frac{5\pi}{4})i+sin(\frac{5\pi}{4})j)}{\sqrt {cos^2(\frac{\pi}{4})+sin^2(\frac{\pi}{4})}\sqrt {cos^2(\frac{5\pi}{4})+sin^2(\frac{5\pi}{4})}}=cos^{-1}\frac{cos(\frac{\pi}{4})~cos(\frac{5\pi}{4})+sin(\frac{\pi}{4})~sin(\frac{5\pi}{4})}{1}=cos^{-1}[cos(\frac{5\pi}{4}-\frac{\pi}{4})]=cos^{-1}[cos~\pi]=\pi$
Remember:
$cos(a-b)=cos~a~cos~b+sin~a~sin~b$
