$\lim _{x \rightarrow 5} \frac{\sqrt{2-2 \cos \left(x^2-12 x+35\right)}}{(x-5)}=\ldots \ldots$

The limit L is evaluated as $x \to 5$ for $\frac{\sqrt{2 - 2\cos(x^2 - 12x + 35)}}{x - 5}$.

Using the identity $1 - \cos A = 2\sin^2\left(\frac{A}{2}\right)$, the numerator simplifies to $\sqrt{4\sin^2\left(\frac{x^2 - 12x + 35}{2}\right)} = 2\left|\sin\left(\frac{(x-5)(x-7)}{2}\right)\right|$.

For the left-hand limit as $x \to 5^-$, note that $(x-5) < 0$ and $(x-7) < 0$, so $(x-5)(x-7) > 0$. Thus the expression inside the sine is positive, and the absolute value is unnecessary: $2\sin\left(\frac{(x-5)(x-7)}{2}\right)$.

Applying the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim_{x \to 5^-} \frac{2\left(\frac{(x-5)(x-7)}{2}\right)}{x-5} = \lim_{x \to 5^-} (x-7) = -2$

Since the right-hand limit yields a different value, the two-sided limit does not exist, but the left-hand limit is $\boxed{-2}$.