= Solution
We use a <closed manifold> of positive <dimension> $n$. The nonnegative <positive Laplace-Beltrami operator> on functions is
$$
\boxed{\Delta=d^*d=-\rho^{-1}\partial_i(\rho g^{ij}\partial_j),\qquad\rho=\sqrt{\det(g_{ij})}.}
$$
It satisfies $\langle\Delta f,f\rangle=\int_M|df|_g^2\,d\mathrm{vol}_g$. Its domain as a <self-adjoint operator> on $L^2(M)$ is $H^2(M)$. The analytic argument in Question 4(d), with $q=0$, gives a smooth <orthonormal eigenbasis> and ordered <eigenvalues> $0=\lambda_0\leq\lambda_1\leq\cdots\to\infty$.
We prove a quantitative counting bound from local <Fourier series>, rather than assume a spectral asymptotic formula. Choose a finite coordinate cover and smooth functions $\chi_\alpha$ compactly supported in its charts with $\sum_\alpha\chi_\alpha^2=1$. This can be obtained by normalizing an ordinary nonnegative <partition of unity> by the square root of the sum of its squares. Embed each chart support in a fixed coordinate <torus>, and extend $v_\alpha=\chi_\alpha f$ by zero there. Uniform comparison of the coordinate and metric norms, and differentiation of the cutoffs, give constants independent of $f$ such that
$$
c\|f\|_2^2\leq\sum_\alpha\|v_\alpha\|_{L^2(\mathbb T^n)}^2,
\qquad
\sum_\alpha\|v_\alpha\|_{H^1(\mathbb T^n)}^2
\leq C\bigl(\|df\|_2^2+\|f\|_2^2\bigr).
$$
Let $P_R$ retain the <Fourier modes> $|k|\leq R$. <Parseval identity> gives
$$
\|(I-P_R)v\|_2^2\leq R^{-2}\|v\|_{H^1}^2\qquad(R\geq1).
$$
The map $f\mapsto(P_Rv_\alpha)_\alpha$ has target <dimension> at most $C_0R^n$, since there are at most $C_nR^n$ lattice points in each cutoff ball.
Let $E_\Lambda$ span all <eigenfunctions> with <eigenvalues> at most $\Lambda$. For $f\in E_\Lambda$, $\|df\|_2^2\leq\Lambda\|f\|_2^2$. If every retained local coefficient vanishes, the preceding estimates imply
$$
c\|f\|_2^2\leq C R^{-2}(1+\Lambda)\|f\|_2^2.
$$
Choose $R=b\sqrt{1+\Lambda}$ with fixed $b$ sufficiently large. Then this is impossible unless $f=0$, so the finite-dimensional coefficient map is injective on $E_\Lambda$. Therefore the <eigenvalue counting function> satisfies
$$
N(\Lambda):=\dim E_\Lambda\leq C_1(1+\Lambda)^{n/2}.
$$
Taking $\Lambda=\lambda_k$ gives $k+1\leq C_1(1+\lambda_k)^{n/2}$, so $\lambda_k\geq c_1(k+1)^{2/n}-1$. For all sufficiently large $k$, this implies $\lambda_k\geq A_0k^{2/n}$.
If $M$ is connected, the zero <eigenvalue> is simple: $\Delta f=0$ implies $df=0$, hence $f$ is constant. Thus every $\lambda_k$ for $k\geq1$ is positive. Taking the minimum of $A_0$ and the finitely many positive ratios $\lambda_k/k^{2/n}$ below the large-$k$ threshold proves
$$
\boxed{\lambda_k\geq Ak^{2/n}\quad(k\geq0),\qquad A>0.}
$$
Connectedness is implicit in the printed claim. If $M$ has $b>1$ connected components, zero has multiplicity $b$, and the claim fails already at $k=1$. The same proof gives $\lambda_k\geq A\max\{k-b+1,0\}^{2/n}$ for all $k$, as well as the universal counting bound above. These finite initial zero modes do not affect part (d).
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