If you allow radicals, then yes to Q1, and depending only on $a(n-1)$. Let:
$$x=\Re( a(n-1))$$
$$y=\Im( a(n-1))$$
$$pos(r) = \frac{1}{2}\left( 1 + \frac{\sqrt{r^2}}{r}\right)$$
Note $pos(r)$ is 1 on positive reals and 0 on negative reals. With this we can enumerate the Gaussian integers in progressively larger diamonds centered on the origin:
\begin{align}
a(0) = &\ 0\\
a(n) = &\ a(n-1) + \\
&(-1 + i)\ pos(+x - 1/2)\ pos(+y -1/2)\ + \\
&(-1 - i)\ pos(-x + 1/2)\ pos(+y -1/2)\ + \\
&(+1 - i)\ pos(-x - 1/2)\ pos(-y + 1/2)\ + \\
&(+1 + i)\ pos(+x + 1/2)\ pos(-y - 1/2)\ + \\
&i\ pos(+x + 1/2)\ \left( pos( y + 1/2) - pos( y - 1/2)\right)
\end{align}
A simple linear modification gives yes to Q2 also.
Even more explicitly:
Mathematica code:\begin{align}
\mathtt{pos[r\_] :=\ } & \mathtt{(1/2) (1 + Sqrt[r^2]/r);}\\
\mathtt{f[z\_] :=\ } & \mathtt{With[{x = Re[z], y = Im[z]}, z +}\\
\mathtt{ }&\mathtt{(-1 + I) pos[+x - 1/2] pos[+y - 1/2] +}\\
\mathtt{ }&\mathtt{(-1 - I) pos[-x + 1/2] pos[+y - 1/2] +}\\
\mathtt{ }&\mathtt{(+1 - I) pos[-x - 1/2] pos[-y + 1/2] +}\\
\mathtt{ }&\mathtt{(+1 + I) pos[+x + 1/2] pos[-y - 1/2] +}\\
\mathtt{ }&\mathtt{I\ pos[+x + 1/2] (pos[y + 1/2] - pos[y - 1/2])];}\\
\mathtt{NestList[} &\mathtt{f,0,13]}\\
\end{align}
Mathematica results, slightly formatted:
$\mathtt{\{0,\\
I,\ -1,\ -I,\ 1,\\
1 + I,\ 2 I,\ -1 + I,\ -2,\ -1 - I,\ -2 I,\ 1 - I,\ 2,\\
2 + I\}}$