= Solution
The integrand is symmetric in $i,j$, so $T$ is a <symmetric tensor>. The sphere and its area element are invariant under rotations, and the integrand transforms covariantly when $\mathbf x,\mathbf y$ are rotated together. For fixed $\mathbf y$, rotations about its direction therefore leave $T$ unchanged. In axes with $\mathbf y$ along the third axis, the mixed transverse-axial entries vanish and the two transverse eigenvalues are equal. Thus it is an <axisymmetric symmetric rank-two tensor>,
$$
T_{ij}=\alpha\delta_{ij}+\beta y_i y_j,
$$
where $\alpha,\beta$ depend on $a,n$ but not on the direction of $\mathbf y$. The restriction $n>-1$ makes the integral convergent near $\mathbf x=\mathbf y$: the integrand is of size distance$^{2n}$ and the local area element has one further radial power.
Take the polar axis along $\mathbf y$. Because $|\mathbf x|=|\mathbf y|=a$,
$$
\boxed{\mathbf y\cdot(\mathbf y-\mathbf x)=a^2(1-\cos\theta),\qquad
|\mathbf y-\mathbf x|^2=2a^2(1-\cos\theta).}
$$
The area element is $a^2\sin\theta\,d\theta\,d\phi$. Hence with $s=1-\cos\theta$,
$$
\begin{aligned}
y_iT_{ij}y_j
&=2\pi a^6(2a^2)^{n-1}\int_0^\pi(1-\cos\theta)^{n+1}\sin\theta\,d\theta\\
&=2\pi a^6(2a^2)^{n-1}\frac{2^{n+2}}{n+2}
=\boxed{\frac{\pi a^2(2a)^{2n+2}}{n+2}}.
\end{aligned}
$$
For the second scalar integral, take the <trace>. It contracts the two chord-vector factors into their squared length:
$$
\begin{aligned}
T_{ii}
&=\int_S|\mathbf y-\mathbf x|^{2n}\,dS\\
&=2\pi a^2(2a^2)^n\int_0^2s^n\,ds
=\frac{\pi(2a)^{2n+2}}{n+1}.
\end{aligned}
$$
Put $K=\pi(2a)^{2n+2}$. The tensor form gives $3\alpha+\beta a^2=K/(n+1)$, while its axial contraction gives $\alpha+\beta a^2=K/(n+2)$. Solving yields the <weighted chord tensor of a sphere> coefficients
$$
\boxed{\alpha=\frac{K}{2(n+1)(n+2)},\qquad
\beta=\frac{K(2n+1)}{2a^2(n+1)(n+2)}.}
$$
Because $a>0$, this is an <isotropic tensor> exactly when $\beta=0$. Therefore \b[$n=-1/2$] is the unique permitted isotropic exponent.
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