If A and B are totally unimodular then is (A 0 0 B)?

Yes, it is true. Here's an outline of my proof, which uses induction. This problem is far harder than your other ones, so I'm not sure how well you'll be able to understand my proof. There may be a simpler one. My final steps dealing with triangular matrices and row reduction require some inspiration or outside knowledge.

Base case: The 2x2 case is true through a brief computation.

Inductive case: Consider
(A 0)
(0 B)
If this matrix is not square, any square submatrix lies in a matrix obtained by deleting some columns and/or rows of A. This is still a block matrix of the above form, so we may inductively suppose that the new matrix is unimodular, so the submatrix in question has determinant 0, 1, or -1.

If this matrix is square, any proper submatrix has determinant 0, 1, or -1 from the same reasoning as above. So, we just need to prove that
(A 0)
(0 B)
has determinant 0, 1, or -1. Consider this matrix over the rational numbers.

(i) First, suppose that A and B are themselves square. Using elementary row operations, we can turn A and B into upper triangular matrices. Applying those same operations to the block matrix, it becomes upper triangular. The determinant of a triangular matrix is the product of the diagonals, and elementary row operations don't change the determinant. It follows that the determinant of the block matrix is det(A)det(B). Since these are square unimodular matrices, it follows that det(A)det(B) = 0, 1, or -1, so the block matrix is indeed unimodular.

(ii) Second, suppose that A is not square (and note that A^T has the same dimensions as B since the block matrix is square). Since det(M) = det(M^T), we may suppose that A has more columns than rows. Let A' be the square matrix obtained from A by cutting off enough of the rightmost columns, and let B' be the square matrix obtained from B by cutting off the topmost rows. Row reduce these and apply the row reductions to the block matrix. The result is upper triangular, though this time there are 0's along the diagonal. Thus the determinant of the block matrix is 0, which is fine--it's still unimodular.