(I) Given $a+2a\cos B=c$, by the Law of Sines, we have

$$\frac{a}{\sin A} = \frac{c}{\sin C}$$

This implies

$$\sin A + 2\sin A\cos B = \sin C$$

From the angle addition formula, we know that $\sin C = \sin(A + B) = \sin A \cos B + \cos A \sin B$.

Now, equating the two expressions for $\sin C$:

$$\sin A + 2\sin A\cos B = \sin A \cos B + \cos A \sin B$$

Simplify to get:

$$\sin A(1 + 2\cos B) = \sin A \cos B + \cos A \sin B$$

Which leads to:

$$\sin A = \cos A \sin B - \sin A \cos B = \sin(B - A)$$

Because $A, B \in (0, \pi)$,

The possible range of $B - A$ is in $(- \pi, \pi)$ and since $A + (B - A) = B \in (0, \pi)$, we have $A + (B - A) \neq \pi$,

Therefore, $A = B - A$, resulting in $\boxed{B = 2A}$.

(II) From part (I), we have $A = \frac{B}{2}$ and $C = \pi - A - B = \pi - \frac{3B}{2}$.

Since $\Delta ABC$ is an acute triangle, the following inequalities must be true:

$$\begin{aligned} 0 &< \frac{B}{2} < \frac{\pi}{2} \\ 0 &< B < \frac{\pi}{2} \\ 0 &< \pi - \frac{3B}{2} < \frac{\pi}{2} \end{aligned}$$

Solving these inequalities, we get $\frac{\pi}{3} < B < \frac{\pi}{2}$.

From $a+2a\cos B=2$, we can find $a$:

$$a = \frac{2}{1 + 2\cos B} \in (1, 2)$$

Therefore, the range of values for $a$ is $\boxed{(1, 2)}$.