What is the angle between <1,3,−8> and <4,1,5>

There are 2 methods we can use to calculate this algebraically, either using the vector cross product or the vector inner product.
The angle between any 2 vectors A and B in any dimensional vector space may be given by the inverse cosine of the Euclidean inner product of the 2 vectors divided by the product of the norms of the 2 vectors.
i.e. ${\displaystyle \cos \theta =\frac{A\cdot B}{\left|\left|A\right|\right|\cdot \left|\left|B\right|\right|}}$
${\displaystyle \therefore \theta ={\cos }^{-1}\left(\frac{\begin{array}{c}(1,3,-8)\cdot (4,1,5)\end{array}}{\left||1,3,-8|\right|\cdot \left||4,1,5|\right|}\right)}$
${\displaystyle ={\cos }^{-1}\left(\frac{4+3-40}{\sqrt{72\sqrt{42}}}\right)}$
${\displaystyle =126,{294}^{\circ }}$