In Gradshteyn's section 17.24 on Fourier transform pairs for spherically symmetric functions, the third entry relates
$\frac{e^{-ar}}{r}$ and $\sqrt{\frac{2}{\pi}}\frac{1}{(a^2 + k^2)^2}$.
I think that there is a mistake on the right hand side, which should be $\sqrt{\frac{2}{\pi}}\frac{1}{a^2 + k^2}$.
My question is if this relation still holds if $a$ is a complex constant.
(1). Yes. There is a mistake on Gradshteyn's book, section 17.24 on p 1120, in the book "Table of integrals, series, and products" 7th. Ed. The right one should be that you pointed out.
(2). Cf. this problem's excellent anser by J.G. Following J.G.'s work, $\tilde{f}(k)=2\pi\int_0^{\pi}{d\theta[\sin{\theta}\int _0^{\infty}{re^{-ar-ikr\cos{\theta}}dr}]}=2\pi\int_0^{\pi}{d\theta[\sin{\theta}\int _0^{\infty}{re^{r\cdot(-a-ik\cos{\theta})}dr}]}\\=2\pi\int_0^{\pi}{d\theta\{\sin{\theta}[\frac{1}{-a-ik\cos{\theta}}\int _0^{\infty}{rde^{r\cdot(-a-ik\cos{\theta})}]}\}}\\=2\pi\int_0^{\pi}{d\theta\{\sin{\theta}\frac{1}{-a-ik\cos{\theta}}[re^{r\cdot(-a-ik\cos{\theta})}|_{r=0}^{\infty}-\int _0^{\infty}{e^{r\cdot(-a-ik\cos{\theta})}dr}]\}}$
,
it is clear that if $a$ is a complex number, to make the integral meaningful, the real part of $a$ should be positive!