1) Call the two numbers x and y. Then you have $\displaystyle \begin{align*} x + y = 12 \end{align*}$ and $\displaystyle \begin{align*} x^2 + y^2 = 80 \end{align*}$. Can you solve these two equations simultaneously for x and y?

2) Call one of the diagonals x and the other y. The diagonals are perpendicular bisectors of the rhombus, so this creates four right angle triangles. The two shorter lengths in one of the triangles are $\displaystyle \begin{align*} \frac{x}{2} \end{align*}$ and $\displaystyle \begin{align*} \frac{y}{2} \end{align*}$. The side length of the rhombus is the hypotenuse, and it is both $\displaystyle \begin{align*} x - 3 \end{align*}$ and $\displaystyle \begin{align*} y - 1 \end{align*}$, so by Pythagoras we have $\displaystyle \begin{align*} \left( \frac{x}{2} \right) ^2 + \left( \frac{y}{2} \right) ^2 = \left( x - 3 \right) ^2 \end{align*}$ and $\displaystyle \begin{align*} \left( \frac{x}{2} \right) ^2 + \left( \frac{y}{2} \right) ^2 = \left( y - 1 \right) ^2 \end{align*}$. Can you solve these equations simultaneously for x and y? If so you can get the side length.

3) The original rectangle has length "l" and width "w", its area is $\displaystyle \begin{align*} 224\,\textrm{cm}^2 \end{align*}$, so $\displaystyle \begin{align*} lw = 224 \end{align*}$. The new rectangle has length $\displaystyle \begin{align*} l - 1 \end{align*}$ and width $\displaystyle \begin{align*} w + 1 \end{align*}$, its area is $\displaystyle \begin{align*} 225\,\textrm{cm}^2 \end{align*}$, so $\displaystyle \begin{align*} \left( l - 1 \right) \left( w + 1 \right) = 225 \end{align*}$. Can you solve these two equations simultaneously for l and w?