I am interested in the solution of the following Fourier transformation
$$\int \frac{1}{|x|^{d+a}}e^{-ixk}d^dx ,$$
considering a general $d$-dimensional system with $a\in\mathbb{R}_+$.
How is this distribution called (to better search for ways to solve it by myself)?
Is there a textbook solution to this problem?
Is there already something on StackExchange regarding this issue?



Similar questions have indeed been asked here now and then, and part of the problem is indeed about keywords to refer to the question.
As commented, some relevant keywords are "homogeneous distribution", also "spherically/rotationally invariant distribution", and such.
An immediate technical obstruction to literal mathematical computation is that, even in the range $\Re(s)<n$, for local integrability, $1/|x|^s$ is not $L^1$ on $\mathbb R^n$. So its Fourier transform cannot be computed by the usual integral. It is also certainly not in $L^2$, so the Plancherel extension of Fourier transform also cannot quite succeed.
Nevertheless, with $\Re(s)<n$ for sure, integration-against $1/|x|^s$ gives a tempered distribution... so it has some kind of Fourier transform, albeit not easily computed by integration.
The idea of a proof is that Fourier transforms of rotationally invariant (tempered) distributions must be rotationally invariant, and homogeneity of degree $s$ (with $1/|x|^s$ as model) is mapped to homogeneity of degree $n-s$ (with $1/|x|^{n-s}$ as model). These two things can be proven in various ways...
Also, it might be good to have in hand a uniqueness result for homogeneous, spherically-symmetric distributions...
Then we "know" that for typical $s$ the FT of (finite part of) $1/|x|^s$ is some multiple (depending on $s,n$) of the (finite part of) $1/|x|^{n-s}$. The anomalous points where this meromorphic family of tempered distribution has poles (in $s$) produces residues which are multiples of $\Delta^\ell \delta$, and Fourier transforms that are integer powers of polynomials $|r|^2$.
The constants can be determined, in terms of the Gamma function, by integrating against Gaussians, since we explicitly know their Fourier transforms (depending on one's choice of normalization of everything...)