Answer
See the detailed answer below.
Work Step by Step
a) We know that the thermal energy of a monatomic gas is given by
$$E_{th,1}=\frac{3}{2}n_1RT$$
and the thermal energy of a diatomic gas is given by
$$E_{th,2}=\frac{5}{2}n_2RT$$
The total thermal energy of the mixture is now given by
$$E_{th,tot}=E_{th,1}+E_{th,2}$$
$$E_{th,tot}=\frac{3}{2}n_1RT+\frac{5}{2}n_2RT$$
$$E_{th,tot}=\frac{RT}{2}(3n_1+5n_2 )$$
Recalling that $E_{th}=nC_{\rm V}T$, so for the mixture, $E_{th}=(n_1+n_2)C_{\rm V}T$
Hence,
$$(n_1+n_2)C_{\rm V}\color{red}{\bf\not} T=\frac{R\color{red}{\bf\not} T}{2}(3n_1+5n_2 )$$
Thus,
$$\boxed{C_{\rm V}=\dfrac{(3n_1+5n_2 )R}{2(n_1+n_2)}}$$
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b) When $n_1=0$, which means that the only gas there is the diatomic.
Hence,
$$C_{\rm V}=\dfrac{(0+5n_2 )R}{2(0+n_2)}=\frac{5}{2}R$$
which is the expected value.
Now let's test it when $n_2=0$ and the only gas there is the monatomic gas.
$$C_{\rm V}=\dfrac{(3n_1+0)R}{2(n_1+0)}=\frac{3}{2}R$$
which is the expected value.