Step 1: Work in the eigenbasis of $B$.
Both Hermitian and anti-Hermitian operators are normal operators, so $B$ always has a complete set of orthonormal eigenvectors. Call its eigenvalues $\lambda_n$, so in this basis $B$ is diagonal with entries $\lambda_n$, and $A = e^{i\theta B}$ is diagonal in the same basis with entries $e^{i\theta \lambda_n}$.
Step 2: Case where $B$ is Hermitian.
A Hermitian operator has only real eigenvalues, so each $\lambda_n$ is real. Each diagonal entry of $A$ is $e^{i\theta \lambda_n}$, a complex number sitting exactly on the unit circle, since $\theta \lambda_n$ is real. A diagonal matrix whose every entry has magnitude 1 is unitary, which confirms statement (A) is TRUE.
Step 3: Determinant in the Hermitian case.
The determinant of a diagonal matrix is the product of its entries, so $\text{Det}(A) = \prod_n e^{i\theta \lambda_n}$. This is a product of unit circle numbers, and such a product always has magnitude 1. So $|\text{Det}(A)| = 1$, confirming statement (C) is TRUE.
Step 4: Case where $B$ is anti-Hermitian.
An anti-Hermitian operator has purely imaginary eigenvalues, so write $\lambda_n = i\mu_n$ with $\mu_n$ real. Each diagonal entry of $A$ becomes $e^{i\theta (i\mu_n)} = e^{-\theta \mu_n}$, a plain real number, not a phase. Since $\theta\mu_n$ need not be zero, this entry does not sit on the unit circle in general, so $A$ is not unitary here, and statement (B) is FALSE.
Step 5: Hermiticity in the anti-Hermitian case.
Every diagonal entry of $A$ works out to the real number $e^{-\theta\mu_n}$, and the basis used is orthonormal, so $A$ is a diagonal matrix with real entries in an orthonormal basis. Such a matrix always equals its own adjoint, so $A$ is Hermitian. This confirms statement (D) is TRUE.
Final Answer:
Diagonalizing $B$ shows directly that (A), (C) and (D) hold while (B) fails.
\[ \boxed{\text{A, C, D}} \]