Part 1 - Let the sides and semiperimeter be $x,y$ and $a$. Then we have: $$xy = 1, x + y = a$$ Solve this system by eliminating y and thus obtaining a quadratic in $x$.
Part 2 - Solve the system of part 1 by using the identity $$\left(\frac{x-y}{2}\right)^2 =\left(\frac{x+y}{2}\right)^2 - xy$$ Not quite sure how to get part 1 but for part 2 I was able to get the following: $$\left(\frac{x-y}{2}\right)^2 = \frac{a^2}{2} -1 $$ $$x - y = ± 2\sqrt{\left(\frac{a}{2}\right)^2 -1} = ± \sqrt{a^2 - 4}$$ $$x = \frac{(x+y) + (x-y)}{2}= \frac{\sqrt{a^2 - 4}}{2}$$ $$y = \frac{(x+y) + (x-y)}{2}= \frac{a±\sqrt{a^2 - 4}}{2}$$
Any confirmation/help with either part is greatly appreciated.