Answer
$ a.\quad$B is an invertible matrix.
$b.\quad X=(B^{-1}+A)^{-1}\cdot A$
Work Step by Step
$(a)$
Multiplying the equation (3) with $X$ from the left, obtaining
$X(A-AX)^{-1}=XX^{-1}B$
$X(A-AX)^{-1}=I\cdot B$
$X(A-AX)^{-1}=B$
The LHS is a product of X, an invertible matrix,
and
$(A-AX)^{-1}$, also an invertible matrix.
By theorem 6, the product of invertible matrices is also an invertible matrix.
Therefore, B is an invertible matrix.
$(b)$
Invert both sides of (3) (both sides are invertible)
$[(A-AX)^{-1}]^{-1}=[X^{-1}B]^{-1}$
$A-AX=B^{-1}\cdot(X^{-1})^{-1}$
$A-AX=B^{-1}\cdot X$
... add $AX$ to both sides,
$A=B^{-1}\cdot X+AX$
... apply the right-distrtibutive property (Th.2)
$A=(B^{-1}+A)X$
$A$ is invertible, so the product $(B^{-1}+A)X$ is invertible
$X $ and $A$ i are invertible, so the factor $(B^{-1}+A)$ must be invertible.
Multiply both sides of the last equation, from the left, with $(B^{-1}+A)^{-1}$.
$(B^{-1}+A)^{-1}\cdot A=(B^{-1}+A)^{-1}\cdot(B^{-1}+A)X$
$(B^{-1}+A)^{-1}\cdot A=I\cdot X$
$(B^{-1}+A)^{-1}\cdot A=X$
