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I have no problem producing a 8 vertices planar graph $G$, but I couldn't figure out how to find complement with $G$ that's also planar.

Is there any systematic way of drawing such graph? Thanks for any help.

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Take an $8$-cycle $v_1v_2v_3v_4v_5v_6v_7v_8v_1$ and add two more edges $v_1v_5$ and $v_2v_6.$ The resulting graph $G$ is clearly planar. I leave it to you to verify that the complement $\overline G$ is also planar. ($\overline G$ is a cube with $6$ additional edges, namely, a diagonal drawn on each face.)

Methodology? It just seemed like a good idea to start with a maximal planar graph; the more edges in the graph, the fewer edges in the complement, and the easier it will be to draw the complement in the plane. So I took a cube (no particular reason, just because it's a nice graph with $8$ vertices) and triangulated each of the faces. The complement turned out to be a cycle plus two diagonals, which was planar and easy to describe.

P.S. To see that $\overline G$ is planar, let $v_1=(0,0),\ v_2=(1,1),\ v_3=(1,2),\ v_4=(1,3),\ v_5=(1,4),\ v_6 (2,4),\ v_7=(2,2),\ v_8=(2,0),$ and draw the edges $v_1v_2,\ v_2v_3,\ v_3v_4,\ v_4v_5,\ v_5v_6,\ v_6v_7,\ v_7v_8,\ v_8v_1,\ v_1v_5,\ v_2v_6$ as straight line segments.

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    "Take an $8$-cycle $v1v2v3v4v5v6v7v8v1v1v2v3v4v5v6v7v8v1$ and add two more edges $v1v5$ and $v2v6$." How is this graph planar? Isn't $v1v5$ and $v2v6$ crossing each other? – 2017-02-01
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    @user59036 if you "pull" $v1v5$ out of the 8-cycle, then it is planar now. – 2017-02-01
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    @user59036 If you draw the $8$-cycle as a ***convex*** polygon, you will not be able to get a ***straight-line*** drawing of $\overline G$ in the plane; you will need one curved edge, as pointed out in the comment by misheekoh. I've edited my answer to show how to draw $\overline G$ in the plane with straight lines. – 2017-02-02