How do I calculate the humidity inside sealed containers containing mixing salts?
As Ivan Neretin pointed out in his comment, there is no simple formula on this matter. Howevr, if you have access to ammonium nitrate ($\ce{NH4NO3}$) or sodium nitrite ($\ce{NaNO2}$), saturated solution of both give 64% relative humidity (RH) at $\pu{25 ^\circ C}$ (Engineering Toolbox).
Interestingly, magnesium nitrate ($\ce{Mg(NO3)2}$) gives exactly 60% RH at $\pu{0 ^\circ C}$, but it was only 54% at $\pu{25 ^\circ C}$. A few salts and their measured RH values in different temperatures are given in the following Table for readers' convenience:
$$
\bf{\text{Relative Humidity over the Saturation Salt Solution (%)}}\\
\begin{array}{c|ccc}
\hline
\text{Salt} &amp; \pu{0 ^\circ C} &amp; \pu{10 ^\circ C} &amp; \pu{20 ^\circ C} &amp; \pu{25 ^\circ C} &amp; \pu{30 ^\circ C} &amp; \pu{40 ^\circ C} &amp; \pu{50 ^\circ C} &amp; \pu{60 ^\circ C}\\
\hline
\ce{NH4NO3} &amp; - &amp; 75 &amp; 67 &amp; 60 &amp; - &amp; - &amp; - &amp; - \\
\ce{(NH4)2SO4} &amp; 82 &amp; 82 &amp; 81 &amp; 81 &amp; 81 &amp; 80 &amp; 79 &amp; - \\
\ce{MgCl2} &amp; 34 &amp; 33 &amp; 33 &amp; 33 &amp; 32 &amp; 32 &amp; 31 &amp; 29 \\
\ce{Mg(NO3)2} &amp;99 &amp; 98 &amp; 98 &amp; 97 &amp; 97 &amp; 96 &amp; 96 &amp; - \\
\ce{KNO3} &amp; 96 &amp; 96 &amp; 95 &amp; 94 &amp; 92 &amp; 89 &amp; 85 &amp; - \\
\ce{KCl} &amp; 89 &amp; 87 &amp; 85 &amp; 84 &amp; 84 &amp; 82 &amp; 81 &amp; 80 \\
\ce{NaCl} &amp; 76 &amp; 76 &amp; 75 &amp; 75 &amp; 75 &amp; 75 &amp; 74 &amp; 75 \\
\ce{NaNO2} &amp; - &amp; - &amp; 65 &amp; 64 &amp; 63 &amp; 61 &amp; - &amp; - \\
\hline
\end{array}
$$