Solution:
Given \( \triangle ABC \), \( AB = 7 \), \( BC = 3 \)

And \( \cos A + 2\cos B + \cos C = 2 \Rightarrow \cos A + \cos C = 2 - 2\cos B \)
\( 2\cos \left( \dfrac{A + C}{2} \right) \cos \left( \dfrac{A - C}{2} \right) = 2(1 - \cos B) \)
As \( A + B + C = 180^{\circ} \), we have
\( \cos \left( \dfrac{180 - B}{2} \right) \cos \left( \dfrac{A - C}{2} \right) = 2\sin^{2} \dfrac{B}{2} \)
\( \sin \dfrac{B}{2} \cos \left( \dfrac{A - C}{2} \right) = 2\sin^{2} \dfrac{B}{2} \Rightarrow \cos \left( \dfrac{A - C}{2} \right) = 2\sin \dfrac{B}{2} \quad \left( \therefore \sin \dfrac{B}{2} \ne 0 \right) \)
Multiplying both sides by \( {\cos\dfrac{B}{2}} \)
we get
\( \cos \dfrac{B}{2} \cos \left( \dfrac{A - C}{2} \right) = 2\sin \dfrac{B}{2} \cos \dfrac{B}{2} \)
\( \Rightarrow \cos \left( \dfrac{180 - (A + C)}{2} \right) \cos \left( \dfrac{A - C}{2} \right) = \sin B \Rightarrow \sin \left( \dfrac{A + C}{2} \right) \cos \left( \dfrac{A - C}{2} \right) = \sin B \)
\( \Rightarrow \dfrac{1}{2} (\sin A + \sin C) = \sin B \Rightarrow \sin A + \sin C = 2\sin B \) ....\( (1) \)
By sine Rule we know that \( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R \) ....\( (2) \)
Where \( a, b, c \) are sides opp. to angles \( A, B, C \) respectively
\( \therefore \) from \( (1) \) and \( (2) \) we get
\( a + c = 2b \)
i.e. \( 3 + 7 = 2b \Rightarrow b = \dfrac{10}{2} = 5 \quad\therefore AC = 5 \)
Now using cosine Rule
We know \(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{5^2+7^2-3^2}{2\times 5\times 7} = \dfrac{65}{70} = \dfrac{13}{14}\)
And \(\cos C = \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{3^2+5^2-7^2}{2\times 3\times 5} = \dfrac{-15}{30} = -1/2\)
Therefore; \(\cos A - \cos C = \dfrac{13}{14} - \left( -\dfrac{1}{2} \right) = \dfrac{13}{14} + \dfrac{1}{2} = \dfrac{20}{14} = \dfrac{10}{7}\)

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