= Solution
We use $\Delta=\operatorname{div}\nabla$, whose <spectrum> on a <compact manifold> is nonpositive; write $P=-\Delta$ for the <positive Laplace-Beltrami operator>. This convention will also make the <heat operator> in Question 3 equal to $\partial_t-\Delta$.
Let $e_1,\ldots,e_d$ be an <orthonormal basis> of $T_pM$, and let $\gamma_i$ be the unit-speed <geodesic> with $\gamma_i(0)=p$ and $\dot\gamma_i(0)=e_i$. Since a <geodesic> has vanishing acceleration for the <Levi-Civita connection>, differentiation twice along it gives the <Riemannian Hessian>:
$$
\left.\frac{d^2}{dt^2}f(\gamma_i(t))\right|_{t=0}
=\operatorname{Hess}_g f(e_i,e_i).
$$
Taking its <metric trace> yields the <geodesic trace formula for the Laplace-Beltrami operator>
$$
\boxed{\Delta f(p)=\sum_{i=1}^d\left.\frac{d^2}{dt^2}f(\gamma_i(t))\right|_{t=0}.}
$$
The positive-sign convention $P$ puts a minus sign in front of this sum.
For the unit <sphere>, take $p\in S^d$ and tangent unit vectors $e_i$ orthogonal to $p$. Its <great circle> <geodesics> are $\gamma_i(t)=p\cos t+e_i\sin t$. For a smooth ambient function $F$ near the <sphere>, the ordinary <chain rule> gives
$$
\left.\frac{d^2}{dt^2}F(\gamma_i(t))\right|_0
=D^2F_p(e_i,e_i)-DF_p(p).
$$
The ambient <orthonormal basis> is $e_1,\ldots,e_d,p$. Thus, writing $F_r(p)=\partial_rF(rp)|_{r=1}$ and similarly for $F_{rr}$,
$$
\boxed{\Delta_{S^d}(F|_{S^d})
=(\widetilde\Delta F)|_{S^d}-F_{rr}|_{r=1}-dF_r|_{r=1}.}
$$
This proves the <ambient restriction formula for the spherical Laplacian> and shows why ambient derivatives normal to the <sphere> must be subtracted. The equivalent formula in <spherical coordinates> is
$$
\widetilde\Delta F=F_{rr}+\frac d rF_r+\frac1{r^2}\Delta_{S^d}(F(r,\cdot)).
$$
Let $\mathcal P_\ell$ be the <vector space> of <homogeneous polynomials> of degree $\ell$ in $d+1$ variables, and $\mathcal H_\ell=\ker\widetilde\Delta\cap\mathcal P_\ell$ its <harmonic polynomials>. If $H\in\mathcal H_\ell$, <homogeneous polynomial> scaling gives $H(r\omega)=r^\ell H(\omega)$. Substitution into the <spherical coordinates> formula gives
$$
\boxed{P_{S^d}(H|_{S^d})=\ell(\ell+d-1)H|_{S^d}.}
$$
The restrictions are the degree-$\ell$ <spherical harmonics>. A <homogeneous polynomial> vanishing on the <sphere> vanishes on every nonzero ray and hence identically, so restriction is injective on $\mathcal H_\ell$.
We prove the <harmonic decomposition of homogeneous polynomials> that both counts these <eigenfunctions> and establishes completeness. For an ambient harmonic <homogeneous polynomial> $H_m$ of degree $m$ and $j\geq1$, the <spherical coordinates> formula gives
$$
\widetilde\Delta(r^{2j}H_m)
=2j(2m+2j+d-1)r^{2j-2}H_m.
$$
For $d\geq1$ its coefficient is nonzero. Inductively assume <homogeneous polynomials> of degree $\ell-2$ have been decomposed into sums of $r^{2j}H_m$. Apply this decomposition to $\widetilde\Delta Q$ for $Q\in\mathcal P_\ell$, and lift each summand by multiplying by $r^2$ and dividing by the displayed nonzero coefficient. The resulting $R\in r^2\mathcal P_{\ell-2}$ satisfies $\widetilde\Delta R=\widetilde\Delta Q$, so $Q-R\in\mathcal H_\ell$. The same coefficient formula shows that no nonzero element of $r^2\mathcal P_{\ell-2}$ can be harmonic. Starting at degrees zero and one therefore proves the direct decomposition
$$
\mathcal P_\ell=\mathcal H_\ell\oplus r^2\mathcal P_{\ell-2}.
$$
Since $\dim\mathcal P_\ell=\binom{\ell+d}{d}$, the <multiplicity> is
$$
\boxed{m_\ell=\binom{\ell+d}{d}-\binom{\ell+d-2}{d},}
$$
where the second term is zero for $\ell<2$.
On $r=1$, iterating the decomposition writes every <polynomial> restriction as a finite sum of <spherical harmonics>. The <Stone-Weierstrass theorem> makes these <polynomial> restrictions dense in $C(S^d)$, since coordinate functions separate points and constants are included; they are consequently dense in $L^2(S^d)$. <Integration by parts> makes $P$ symmetric on smooth functions and gives orthogonality between distinct <eigenvalues>. Normalizing within each finite-dimensional <eigenspace> gives a complete <orthonormal eigenbasis> $Y_{\ell a}$. To check the full operator <spectrum>, let $c_{\ell a}=\langle f,Y_{\ell a}\rangle$. For smooth $f$, <integration by parts> gives the coefficients of $Pf$ as $\lambda_\ell c_{\ell a}$. Therefore finite harmonic partial sums approximate both $f$ and $Pf$ in $L^2$, hence approximate in the <graph norm>. Conversely every coefficient sequence with $\sum_{\ell,a}\lambda_\ell^2|c_{\ell a}|^2<\infty$ is the <graph norm> limit of its finite harmonic sums. Thus the closure of $P$ has exactly this <operator domain> and is the real diagonal <self-adjoint operator> with entries $\lambda_\ell=\ell(\ell+d-1)$. If $z$ is not one of these entries, the inverse has entries $(\lambda_\ell-z)^{-1}$ and is bounded because $\lambda_\ell\to\infty$. This proves that there are no missing spectral values. The <spectrum of the Laplacian on a sphere> is
$$
\boxed{\sigma(P)=\{\ell(\ell+d-1):\ell=0,1,2,\ldots\},}
$$
with the above <multiplicities>; for $\Delta$, negate every <eigenvalue>. For example $S^1$ has <eigenvalues> $\ell^2$, with <multiplicities> one at zero and two for $\ell\geq1$, and $S^2$ has <multiplicity> $2\ell+1$. The zero-dimensional sphere consists of two isolated points and has only the zero <eigenvalue>, with <multiplicity> two.
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