To solve this problem, we need to analyze the implications of the conditions given and how they relate to the properties of a linear transformation and its range and null space.

Given: Let \( V(F) \) be a finite-dimensional vector space and \( T: V \to V \) be a linear transformation. Let \( R(T) \) denote the range of \( T \) and \( N(T) \) denote the null space of \( T \). The condition is \( \text{rank}(T) = \text{rank}(T^2) \).

The rank-nullity theorem states:

\text{dim}(V) = \text{rank}(T) + \text{nullity}(T)

The condition \( \text{rank}(T) = \text{rank}(T^2) \) implies that the range of \( T \) and \( T^2 \) are closely linked. Specifically, this means the image under \( T \) is stabilized, and applying \( T^2 \) does not further reduce the dimensionality.

1. Option A: \( N(T) = R(T) \)

   This situation is generally not true as it would imply that the null space and the range are the same, which is an unusual situation and not implied by the given condition. Therefore, Option A is inconsistent with the typical properties of linear transformations.

2. Option B: \( N(T) = N(T^2) \)

   This is true under the given condition. Since \( \text{rank}(T) = \text{rank}(T^2) \), any element that becomes zero after applying \( T^2 \) is already zero when \( T \) is applied. Hence, the null spaces coincide.

3. Option C: \( N(T) \cap R(T) = \{0\} \)

   This implies that the intersection of the null space and the range is only the zero vector. Since \( \text{rank}(T) = \text{rank}(T^2) \), \( T \) is injective on its range, and elements in the null space cannot map to a non-zero element of the range through \( T \).

4. Option D: \( R(T) = R(T^2) \)

   Given \( \text{rank}(T) = \text{rank}(T^2) \), the range of \( T \) and \( T^2 \) must be the same. This shows stability, meaning applying \( T \) again does not expand the range. Thus, this option is correct.

Consequently, the correct options are B, C, and D.