= Solution
The <chemical potential> is $\mu=(\partial E/\partial N)_{S,V}$, equivalently $(\partial F/\partial N)_{T,V}$: it is the free-energy cost of adding a particle at fixed temperature and volume. With units $k_B=1$, a large reservoir has entropy $S_R(E_{\rm tot}-E,N_{\rm tot}-N)=S_R^0-E/T+\mu N/T$ to first order. Since the number of compatible reservoir states is proportional to its exponential, the <grand canonical ensemble> assigns
$$
\boxed{\Pr(\text{state }j,N)=\Xi^{-1}e^{-(E_{j,N}-\mu N)/T},\quad\Xi=\sum_{N,j}e^{-(E_{j,N}-\mu N)/T}.}
$$
For independent <fermions>, a one-particle state of energy $\varepsilon$ has occupation zero or one. Its partition factor is $1+e^{-(\varepsilon-\mu)/T}$, so the mean occupation is the <Fermi-Dirac distribution> $[e^{(\varepsilon-\mu)/T}+1]^{-1}$. Multiplication by the given <density of states> yields $dN=C\varepsilon^{1/2}d\varepsilon/[e^{(\varepsilon-\mu)/T}+1]$.
At zero temperature the occupation is a step function, filled below $\mu$ and empty above it, so $\mu=\varepsilon_F$ and $N=(2C/3)\varepsilon_F^{3/2}$. At fixed particle number, the low-temperature shift of $\mu$ is $O(T^2/\varepsilon_F)$ and does not affect the leading estimate. In the narrow interval of width $T$ above $\varepsilon_F$, replace the density of states by $C\sqrt{\varepsilon_F}$ and put $u=(\varepsilon-\varepsilon_F)/T$. Then
$$
\boxed{N_{>\varepsilon_F}\sim C\sqrt{\varepsilon_F}\,T\int_0^\infty\frac{du}{e^u+1}=C\sqrt{\varepsilon_F}\,T\log2.}
$$
Equivalently, the fraction above the <Fermi energy> is asymptotic to $(3\log2/2)T/\varepsilon_F$.
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