Given,

$AB=7, BC=3$.

$\Rightarrow c=7, a=3$.

Apply cosine rule in the given equation $\cos A+2\cos B+\cos C=2$, we get

$\Rightarrow \left(\frac{b^2+c^2-a^2}{2bc}\right)+2\left(\frac{c^2+a^2-b^2}{2ac}\right)+\left(\frac{a^2+b^2-c^2}{2ab}\right)=2$

$\Rightarrow \left(\frac{b^2+49-9}{14b}\right)+\left(\frac{49+9-b^2}{21}\right)+\left(\frac{9+b^2-49}{6b}\right)=2$

$\Rightarrow \left(\frac{b^2+40}{14b}\right)+\left(\frac{58-b^2}{21}\right)+\left(\frac{b^2-40}{6b}\right)=2$

$\Rightarrow b^3-5b^2-16b+80=0$

$\Rightarrow \left(b-4\right)\left(b+4\right)\left(b-5\right)=0$

$\Rightarrow b=-4,4,5$.

But $b\ne -4$.

Also, if $b=4$ triangle cannot be constructed.

That means $b=5$.

$\Rightarrow \cos A-\cos C=\left(\frac{b^2+c^2-a^2}{2bc}\right)-\left(\frac{a^2+b^2-c^2}{2ab}\right)$

$\Rightarrow \cos A-\cos C=\left(\frac{5^2+7^2-3^2}{2\left(5\times 7\right)}\right)-\left(\frac{3^2+5^2-7^2}{2\left(3\times 5\right)}\right)$

$\Rightarrow \cos A-\cos C=\left(\frac{65}{70}\right)-\left(\frac{-15}{30}\right)=\frac{195+105}{210}=\frac{300}{210}$

$\Rightarrow \cos A-\cos C=\frac{10}{7}$