Proving matrix properties: 
Prove:
(i) $A(I+BA)^{-1}=(I+AB)^{-1}A$
(ii) $(I+AB)^{-1}=I-A(I+BA)^{-1}B$

(i)
Consider $A(I+BA)=(A+ABA)=(I+AB)A$
Taking inverse on both sides (invert)
$[A(I+BA)]^{-1}=[(I+AB)A]^{-1}$
$(I+BA)^{-1}A^{-1}=A^{-1}(I+BA)^{-1}$
Pre-multiply by A
A(I+BA)^{-1}A^{-1}=A A^{-1}(I+AB)^{-1}$
Post multiply by A
$A(I+BA)^{-1}A^{-1}A=(I+AB)^{-1}A$
$A(I+BA)^{-1}=(I+AB)^{-1}A$, HENCE SHOWN.
(ii)
We start with
$(I+AB)(I+AB)^{-1}=I$
Expand brackets:
$(I+AB)^{-1}+AB(I+AB)^{-1}=I$
$(I+AB)^{-1}+AB[B^{-1}(B+BAB)]^{-1}=I$
$(I+AB)^{-1}+AB[B^{-1}(I+BA)B]^{-1}=I$
Careful!
$(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$
So,$(I+AB)^{-1}+AB[B^{-1}(I+BA)B]^{-1}=I$ becomes
$(I+AB)^{-1}+AB[B^{-1}(I+BA)^{-1}B]=I$
$(I+AB)^{-1}+ABB^{-1}(I+BA)^{-1}B=I$
$(I+AB)^{-1}+A(I+BA)^{-1}B=I$
$(I+AB)^{-1}=I-A(I+BA)^{-1}B$. HENCE SHOWN
 A: For (i) we can multiply both sides with $(I+BA)$ (from the right) to obtain that (i) holds iff \begin{align}A(I+BA)^{-1}(I+BA)&=(I+AB)^{-1}A(I+BA)\\[0.2cm]\iff A&=(I+AB)^{-1}(A+ABA)\\[0.2cm]\iff A&=(I+AB)^{-1}(I+AB)A\\[0.2cm]\iff A&=A\end{align} which is true.
For (ii) observe that the RHS can be written (of course in view of the LHS of (i)) as \begin{align}I-A(I+BA)^{-1}B&\overset{(i)}=I-(I+AB)^{-1}AB\\[0.2cm]&=(I+AB)^{-1}(I+AB)-(I+AB)^{-1}AB\\[0.2cm]&=(I+AB)^{-1}(I+AB-AB)\\[0.2cm]&=(I+AB)^{-1}\end{align}

Equivalently, for (i), you can directly write \begin{align}A&=(I+AB)^{-1}(I+AB)A=(I+AB)^{-1}(A+ABA)=(I+AB)^{-1}A(I+BA)\end{align} Hence multiplying both sides from the right with $(I+BA)^{-1}$ gives you $$A(I+BA)^{-1}=(I+AB)^{-1}A(I+BA)(I+BA)^{-1}=(I+AB)^{-1}A$$ which is what you wanted to prove.
A: For the first one, multiply by $(I-AB)$ and $(I-BA)$ to remove the inverses, expand and compare. Should be easy from there. For number two, use number 1 to change the right-hand side up a bit, then multiply to get rid of the inverses, and you should be able to compare the two sides easily.
