Sorry for the crappy microsoft paint pictures.
The longer chord is 1/3 of the way from the bottom of the circle, so it's 2/3 of a radius up from the bottom. Let $R$ be the radius of the circle. Then we can label the vertical segments $2R/3$ and $R/3$, and we can label the diagonal radius $R$. The chord has length $4$ and is bisected by the vertical radius, so we can label each half with length $2$. Now apply the Pythagorean theorem to the right triangle with sides $2$, $R/3$ and $R$:
\begin{align*}
\left(\frac{R}{3}\right)^2 + 2^2 &= R^2 \\
\frac{R^2}{9}+4 &= R^2 \\
R^2 + 36 &= 9R^2 \quad \text{(multiply through by 9)}\\
36 &= 8 R^2 \\
\frac{9}{2} &= R^2\\
\frac{3}{\sqrt{2}} &= R
\end{align*}
so finally the diameter is $D=2R=2\frac{3}{\sqrt{2}}= 3\sqrt{2} \approx4.2426' \approx 4' 2 \frac{7}{8}''.$
For the second chord, it is $10\%$ of the way from the bottom, so $1/5$ of a radius from the bottom. So we get the vertical segment labeled $4R/5$. Again there is a diagonal radius of length $R$, and we have $L/2$, half the length of the shorter chord. Applying the Pythagorean theorem:
\begin{align*}
\left(\frac{4R}{5}\right)^2 + \left(\frac{L}{2}\right)^2 &= R^2 \\
\frac{16R^2}{25}+ \frac{L^2}{4} &= \frac{25 R^2}{25} \\
\frac{L^2}{4} &= \frac{25R^2-16R^2}{25}\\
\frac{L^2}{4}&= \frac{9R^2}{25}\\
\frac{L}{2} &= \frac{3R}{5}\\
L&= \frac{6R}{5} = \frac{9\sqrt{2}}{5}
\end{align*}
i.e., $L \approx 2.5456' \approx 2' 6 \frac{1}{2}''.$
