Solve in Reals
$$\begin{cases} \cos x+\cos y=\frac{1}{2} \\ \tan x +\tan y=2\\ \text{I tried:}\\ \frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}=2\rightarrow \sin x \cos y+\sin y\cos x=2\cos x\cos y\rightarrow \\ \frac{1}{2}[\sin(x-y)+\sin(x+y)] +\frac{1}{2}[\sin(x+y)-\sin(x-y)]=2\cos x\cos y\\ \sin(x-y)+\sin(x+y)+\sin(x+y)-\sin(x-y)=4\cos x\cos y \rightarrow\\ 2\sin(x-y) = 4\cos x\cos y\\ \sin(x-y)=2\cos x\cos y \rightarrow \sin x\cos y-\cos x\sin y=2 \cos x\cos y \\ \sin x\cos y-\cos x\sin y-2\cos x\cos y=0 \rightarrow \sin x\cos y-\cos x(\sin y+2\cos y)=0??? \end{cases}$$
\begin{align} \cos x+\cos y=\tfrac{1}{2} \tag{1}\label{1} ,\\ \tan x +\tan y=2 \tag{2}\label{2} . \end{align}
Subtracting $(\tan y +1)$ from \eqref{2} and squaring, we get
\begin{align} (\tan x-1)^2 &= (1-\tan y)^2 ,\\ \tan^2x+1-2\tan x &= \tan^2y+1-2\tan y \tag{3}\label{3} ,\\ \frac{1}{\cos^2x}-2\tan x &= \frac{1}{\cos^2y}-2\tan y \tag{4}\label{4} . \end{align}
Substitution of $\tan y=2-\tan x$ into \eqref{4} and rearrangement results in
\begin{align} \frac{1}{\cos^2x} -\frac{1}{\cos^2y} +4&=4\tan x=4\,\frac{\sin x}{\cos x} \tag{5}\label{5} .\\ \end{align}
Substitution of $\cos y=\tfrac12-\cos x$ into \eqref{5} results in
\begin{align} \frac{1}{\cos^2x} -\frac{1}{(\tfrac12-\cos x)^2} +4&=4\,\frac{\sin x}{\cos x} \tag{6}\label{6} ,\\ \frac{1}{\cos x} -\frac{\cos x}{(\tfrac12-\cos x)^2} +4\,\cos x&=4\,\sin x \tag{7}\label{7} . \end{align}
After squaring \eqref{7} we have
\begin{align} \frac{ (1-4\,\cos x+4\,\cos^2 x-16\,\cos^3 x +16\,\cos^4 x)^2}{ \cos^2x\,(1-2\,\cos x)^4} &= 16\,\sin^2x=16\,(1-\cos^2x) \tag{8}\label{8} . \end{align}
Assuming that $\cos x\ne0$, $1-\cos^2x\ne0$, we arrive at equation in $t=\cos x$ \begin{align} 512\,t^8-1024\,t^7+512\,t^6 +128\,t^5-192\,t^4+64\,t^3+8\,t^2-8\,t+1 &=0 \tag{9}\label{9} .\\ \end{align}
Luckily, \eqref{9} can be factored into \begin{align} (32\,t^4-32\,t^3+4\,t-2-\sqrt2) (32\,t^4-32\,t^3+4\,t-2+\sqrt2) &=0 \tag{10}\label{10} , \end{align}
which has six real roots: \begin{align} t_{1}&=\tfrac14+\tfrac14\,\sqrt{3+2\,\sqrt{5+2\,\sqrt2}},\\ t_{2}&=\tfrac14-\tfrac14\,\sqrt{3+2\,\sqrt{5+2\,\sqrt2}},\\ t_{3}&=\tfrac14+\tfrac14\,\sqrt{3+2\,\sqrt{5-2\,\sqrt2}},\\ t_{4}&=\tfrac14-\tfrac14\,\sqrt{3+2\,\sqrt{5-2\,\sqrt2}},\\ t_{5}&=\tfrac14+\tfrac14\,\sqrt{3-2\,\sqrt{5-2\,\sqrt2}},\\ t_{6}&=\tfrac14-\tfrac14\,\sqrt{3-2\,\sqrt{5-2\,\sqrt2}} . \end{align}
Equations \eqref{1},\eqref{2} are symmetric in $x,y$, and indeed,
\begin{align} t_2&=\tfrac12-t_1 ,\\ t_4&=\tfrac12-t_3 ,\\ t_6&=\tfrac12-t_5 , \end{align}
so, basically, we can consider just three of them.
Next, \begin{align} \cos x&=t_i,\quad \cos y=\tfrac12-t_i ,\\ x&=\pm \arccos (t_i)+2\pi\,n,\quad y=\pm \arccos(\tfrac12-t_i)+2\pi\,m , \end{align}
so we need to check which combinations of the signs satisfy \eqref{2}.
And the three suitable pairs of $x,y$ for $n=m=0$ happens to be \begin{align} (\arccos t_1,\ -\arccos(\tfrac12-t_i)) &\approx(0.1848292031,\ -2.074837017) ,\\ (-\arccos t_3,\ -\arccos(\tfrac12-t_3)) &\approx(-0.5361638202,\ -1.938715548) ,\\ (-\arccos t_5,\ \arccos(\tfrac12-t_5)) &\approx(-1.258318311,\ 1.377003189) . \end{align}