= Solution
The unit sphere and the kernel $e^{-c|y-x|^2}$ are invariant under every <rotation in three dimensions> fixing $y$. Therefore $T(y)$ has the same <axisymmetry>, so part (a) gives
$$
T_{ij}=\alpha\delta_{ij}+\beta y_iy_j.
$$
Choose <spherical polar coordinates> with polar axis $y$ and put $u=\cos\theta=x\mathbin\cdot y$. Since $|x|=|y|=1$,
$$
|y-x|^2=2(1-u),
\qquad dA=2\pi\,du.
$$
The <tensor contraction> giving the trace is
$$
T_{kk}=\int_S|x|^2e^{-c|y-x|^2}\,dA
=2\pi\int_{-1}^1e^{-2c(1-u)}du
=\boxed{\frac{\pi}{c}(1-e^{-4c})}.
$$
Similarly,
$$
T_{ij}y_iy_j
=2\pi\int_{-1}^1u^2e^{-2c(1-u)}du
=\boxed{\frac{\pi e^{-2c}}{c^3}
\left[(2c^2+1)\sinh(2c)-2c\cosh(2c)\right]}.
$$
At $c=0$ these expressions have the continuous limiting values $4\pi$ and $4\pi/3$. Writing them as $A=T_{kk}=3\alpha+\beta$ and $B=T_{ij}y_iy_j=\alpha+\beta$ gives
$$
\boxed{\alpha=\frac{A-B}{2},\qquad \beta=\frac{3B-A}{2}}.
$$
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