I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.
You cannot generally just average both pressures. As $pV=nRT$, then ( for $\mathrm{T}$ and $\mathrm{V}$ constant ) :
$V \cdot \mathrm{d}p=\mathrm{d}n \cdot RT$
$\frac{\mathrm{d}p}{\mathrm{d}n}=\frac{RT}V$.
So pressure averaging would work only in the special case if $\frac{T_1}{V_1}=\frac{T_2}{V_2}$. Therefore, a weighted average must be used.
I suppose the scenario is vaguely defined. It can be the polytropic expansion, or it may be the "locally isothermal" expansion. But for the former, it has like infinite number of possible solutions, depending on the rate of the transfer and of heat exchange.
I consider below the latter, i.e. isothermal conditions in each of containers.
We can consider the ideal behaviour, where
$$p=\frac{nRT}V$$
For equality of pressures, if $\Delta n$ is the moved molar amount of xenon from the container 2 to the container 1, the following equality must be true:
$$p=\frac{(n_1 + \Delta n)RT_1}{V_1}=\frac{(n_2 - \Delta n)RT_2}{V_2}$$
From that, we can easily calculate $\Delta n$:
$$(n_1 + \Delta n)RT_1V_2=(n_2 - \Delta n)RT_2V_1$$
$$\Delta n \cdot R(T_1V_2 + T_2V_1)=n_2RT_2V_1 - n_1RT_1V_2$$
$$\Delta n=\frac{n_2T_2V_1 - n_1T_1V_2}{T_1V_2 + T_2V_1}$$
Now you plug in the numbers here in the final formula, instead of working with literal numbers all the way:
In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?
$$\Delta n=\frac{(\pu{100 mol})(\pu{1573.15 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573.15 K})(\pu{4.000 m3})}{(\pu{573.15 K})(\pu{4.000 m3}) + (\pu{1573.15 K})(\pu{2.780 m3})}$$
In the sense below, if to be precise, we should round the temperature values for presented formula, because temperature was not known in 2 decimals accuracy ( otherwise there would be $\pu{300.00 ^{\circ}C}$, resp. $\pu{1300.00 ^{\circ}C}$), internally still computing with non rounded values.
$$\Delta n=\frac{(\pu{100 mol})(\pu{1573 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573 K})(\pu{4.000 m3})}{(\pu{573 K})(\pu{4.000 m3}) + (\pu{1573 K})(\pu{2.780 m3})} \approx \pu{-3.18 mol}$$
The negative value means the $\pu{3.18 mol}$ of xenon is to be moved from the first container to the second one, as we have formally assumed the opposite direction.
Internal full calculator precision of computation still makes good sense, to avoid artefacts of rounding errors. But for presentation of final or intermediate values of quantities from the real world, the values have to reflect error propagation rules. As input data and constants have limited accuracy.
