Answer
a) $4$
b) $1$
c) $16$
Work Step by Step
a) We know that the rms speed is given by
$$v_{\rm rms}=\sqrt{\dfrac{k_BT}{M}}$$
Hence,
$$\dfrac{(v_{\rm rms})_{H_2}}{(v_{\rm rms})_{O_2}}=\dfrac{\sqrt{\dfrac{k_BT_{H_2}}{M_{H_2}}}}{\sqrt{\dfrac{k_BT_{O_2}}{M_{O_2}}}}$$
We know that the mixture will have one final temperature after a long time. So, $T_{H_2}=T_{O_2}$;
$$\dfrac{(v_{\rm rms})_{H_2}}{(v_{\rm rms})_{O_2}}= \sqrt{\dfrac{M_{O_2}}{M_{H_2}}}= \sqrt{\dfrac{2M_{O}}{2M_{H}}}$$
Plugging the known;
$$\dfrac{(v_{\rm rms})_{H_2}}{(v_{\rm rms})_{O_2}}= \sqrt{\dfrac{2(16\;\rm u)}{2(1\;\rm u)}}$$
$$\dfrac{(v_{\rm rms})_{H_2}}{(v_{\rm rms})_{O_2}}=\color{red}{\bf 4}$$
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b) We know that the average kinetic energy of a molecule is given by
$$K=\frac{3}{2}k_BT$$
Hence,
$$\dfrac{K_{H_2}}{K_{O_2}}=\dfrac{\frac{3}{2}k_BT_{H_2}}{\frac{3}{2}k_BT_{O_2}}$$
and since the two temperatures are equal,
$$\dfrac{K_{H_2}}{K_{O_2}}=\color{red}{\bf1}$$
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c) We know that the thermal energy of an ideal diatomic gas is given by
$$E_{th}=\frac{5}{2}nRT$$
And we know that the number of moles is given by
$$n=\dfrac{m}{M}$$
where $m$ is the mass of the sample and $M$ is the atomic mass of the gas molecule or atom.
Hence,
$$E_{th}=\frac{5}{2}\dfrac{m}{M}RT$$
Thus,
$$\dfrac{(E_{th})_{H_2}}{(E_{th})_{O_2}}= \dfrac{\frac{5}{2}\dfrac{m_{H_2}}{M_{H_2}}RT_{H_2}}{\frac{5}{2}\dfrac{m_{O_2}}{M_{O_2}}RT_{O_2}}$$
We know that $T_{H_2}=T_{O_2}$, and that $m_{H_2}=_{O_2}$,
$$\dfrac{(E_{th})_{H_2}}{(E_{th})_{O_2}}= \dfrac{M_{O_2}}{M_{H_2}}= \dfrac{2M_{O}}{2M_{H}}$$
Plugging the known;
$$\dfrac{(E_{th})_{H_2}}{(E_{th})_{O_2}} = \dfrac{2(16\;\rm u)}{2(1\;\rm u)} $$
$$\dfrac{(E_{th})_{H_2}}{(E_{th})_{O_2}} =\color{red}{\bf16}$$
