Answer
See the detailed answer below.
Work Step by Step
First, we need to find the number of moles, or the number of molecules, of each gas.
We need to use the ideal gas law of
$$PV=nRT=Nk_BT$$
Hence,
$$n=\dfrac{PV}{RT}$$
Thus,
$$n_{He}=\dfrac{P_{He}V_{He}}{RT_{He}}=\dfrac{(2\times 1.013\times 10^5)(100\times 10^{-6})}{(8.31)(100+273)}$$
$$n_{He}=\bf 6.536\times 10^{-3} \;\rm mol$$
$$n_{Ar}=\dfrac{P_{Ar}V_{Ar}}{RT_{Ar}}=\dfrac{(4\times 1.013\times 10^5)(200\times 10^{-6})}{(8.31)(400+273)}$$
$$n_{Ar}=\bf 1.449\times 10^{-2} \;\rm mol$$
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a) The thermal energy of a monatomic gas is given by
$$E_{th}=\frac{3}{2}nRT$$
Hence,
$$(E_{th})_{i,He}=\frac{3}{2}(6.536\times 10^{-3})(8.31)(100+273)$$
$$(E_{th})_{i,He}=\color{red}{\bf 30.4}\;\rm J$$
$$(E_{th})_{i,Ar}=\frac{3}{2}(1.449\times 10^{-2})(8.31)(400+273)$$
$$(E_{th})_{i,Ar}=\color{red}{\bf 122}\;\rm J$$
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d) Let's assume that the two gases are well isolated from surroundings which means that any heat loss from one gas will be absorbed by the other.
$$Q_{He}+Q_{Ar}=0$$
$$Q_{He}=-Q_{Ar}\tag 1$$
Recalling that $\Delta E_{th}=Q+W$
And since the volume of both gases remains constant during this thermal connection, the work done on each gas of them is zero.
Hence, $Q=\Delta E_{th}$.
Plugging into (1);
$$(\Delta E_{th})_{He}=-(\Delta E_{th})_{Ar} $$
$$ \color{red}{\bf\not} \frac{3}{2}n_{He} \color{red}{\bf\not} R(T_f-T_{He})=- \color{red}{\bf\not} \frac{3}{2}n_{Ar} \color{red}{\bf\not} R(T_f-T_{Ar})$$
The two gases, finally, must have the same final temperature.
$$ n_{He} (T_f-T_{He})=- n_{Ar} (T_f-T_{Ar})$$
Thus,
$$ n_{He} T_f-n_{He}T_{He} =- n_{Ar} T_f+ n_{Ar}T_{Ar} $$
$$ n_{He} T_f+ n_{Ar} T_f=n_{He}T_{He} + n_{Ar}T_{Ar} $$
So,
$$T_f=\dfrac{n_{He}T_{He} + n_{Ar}T_{Ar}}{n_{He} + n_{Ar}}$$
Plugging the known;
$$T_f=\dfrac{ (6.536\times 10^{-3})(100+273)+(1.449\times 10^{-2})(400+273)}{(6.536\times 10^{-3}) + (1.449\times 10^{-2})}$$
$$T_f=\color{red}{\bf 580}\;\rm K=\color{red}{\bf 307}^\circ C$$
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b) Now it is easy to find the final thermal energy of each gas:
$$(E_{th})_{f,He}=\frac{3}{2}(6.536\times 10^{-3})(8.31)(307+273)$$
$$(E_{th})_{f,He}=\color{red}{\bf 47.3}\;\rm J$$
$$(E_{th})_{f,Ar}=\frac{3}{2}(1.449\times 10^{-2})(8.31)(307+273)$$
$$(E_{th})_{f,Ar}=\color{red}{\bf 105}\;\rm J$$
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c)
The final temperature of the system is greater than the initial temperature of the Helium gas and is less than the initial temperature of the Argon gas.
Hence, the heat is transferred from Argon to Helium.
This much heat can be found by finding the change in thermal energy of any gas of them.
Hence,
$$Q=(\Delta E_{th})_{He}=47.3-30.4=\color{red}{\bf 16.9}\;\rm J$$
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e)
The final pressure is given by the ideal gas law;
$$P_fV=nRT_f$$
Hence,
$$P_f=\dfrac{nRT_f}{V}$$
Thus,
$$(P_f)_{He}=\dfrac{n_{He}RT_f}{V}=\dfrac{(6.536\times 10^{-3})(8.31)(580)}{(100\times 10^{-6})}$$
$$(P_f)_{He}=\color{red}{\bf 3.15\times 10^5}\;\rm Pa=\color{red}{\bf 3.1}\;\rm atm$$
And,
$$(P_f)_{Ar}=\dfrac{n_{Ar}RT_f}{V}=\dfrac{(1.449\times 10^{-2} )(8.31)(580)}{(200\times 10^{-6})}$$
$$(P_f)_{He}=\color{red}{\bf 3.5\times 10^5}\;\rm Pa=\color{red}{\bf 3.45}\;\rm atm$$
