1. Identify the matrix form

Given

$$A=\frac12\begin{bmatrix}1&\sqrt3\\-\sqrt3&1\end{bmatrix} =\begin{bmatrix}\frac12&\frac{\sqrt3}{2}\\-\frac{\sqrt3}{2}&\frac12\end{bmatrix}.$$

Compare this with the standard rotation matrix

$$R(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}.$$

Our matrix is

$$A=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}=R(-\theta)$$

with

$$\cos\theta=\frac12,\qquad \sin\theta=\frac{\sqrt3}{2}.$$

So,

$$\theta=\frac\pi3.$$

Hence,

$$A=R\left(-\frac\pi3\right).$$

2. Find powers of $A$

For rotation matrices,

$$R(\alpha)^n=R(n\alpha).$$

Therefore,

$$A^n=R\left(-\frac{n\pi}{3}\right).$$

So,

$$A^{30}=R\left(-\frac{30\pi}{3}\right)=R(-10\pi)=I$$

because rotation by any multiple of $2\pi$ is the identity.

Also,

$$A^{25}=R\left(-\frac{25\pi}{3}\right).$$

Reduce the angle modulo $2\pi=\frac{6\pi}{3}$:

$$-\frac{25\pi}{3}\equiv -\frac{\pi}{3}\pmod{2\pi}.$$

Thus,

$$A^{25}=R\left(-\frac\pi3\right)=A.$$

Hence,

$$A^{30}=I,\qquad A^{25}=A.$$

3. Check each option

Option A

$$A^{30}-A^{25}=I-A.$$

This is not equal to $2I$.

So, A is false.

Option B

$$A^{30}+A^{25}-A=I+A-A=I.$$

So, B is true.

Option C

$$A^{30}=A^{25}\iff I=A,$$

which is false since $A\neq I$.

So, C is false.

Option D

$$A^{30}+A^{25}+A=I+A+A=I+2A,$$

which is not equal to $I$.

So, D is false.

4. Final answer

The correct option is

$$\boxed{\text{B}}$$