a) If a square matrix \(A\) satisfies \(A^{17}=0\), then it follows that \(I-A\) is invertible.

To prove this, we will use the formula for the inverse of a matrix expressed as a sum:

\[

(I-A)^{-1}=\sum_{k=0}^{\infty} A^k

\]

Since \(A^{17}=0\), it follows that all higher powers of \(A\) vanish as well. So, the sum is finite and well-defined, and the inverse exists.

b) Not necessarily. If \(B\) is invertible, it doesn't necessarily mean that \(B-A\) is invertible.

For a counterexample, consider the matrix \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\). Both matrices are invertible, but \(B-A=\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]\) is not invertible.