Question 4.23 asks you to prove the converse of the Pythagorean Theorem.  So if you have a triangle $ABC$, whose side lengths satisfy the equation $a^2+b^2=c^2$, you want to prove that $\angle C=90^\circ$.  The idea to prove this result is to create a "helping" triangle $A'B'C'$, with $\angle C'=90^\circ$, and $a'=a$, $b'=b$.  Can you show that these two triangles are congruent?

For Question 4.25, if you are considering using Heron's formula, go ahead and try it out, and see if it works.