First, we set some notations.

a) Consider a side parallel to the $x$ axis. Two vertices of this side are vertices of two angles which are in the $xy$ or $zx$ plane. On the other hand, each angle parallel to the $xy$ plane has a side parallel to the $y$ axis and a side parallel to the $x$ axis. Similar arguments for other directions imply
$$Y + Z = 2A, \quad Z + X = 2B, \quad X + Y = 2C$$
Therefore, $X$, $Y$ and $Z$ have the same parity.

b) According to the part (a), we have two cases for a lattice hexagon which is not planar.
$$\bullet X = Y = Z = 2.$$
$$\bullet \{X, Y, Z\} = \{0, 2, 4\}.$$
In the first case, we can divide the polygon into two rectangles. Now consider the lines $l$ and $l'$ perpendicular to the plane of each rectangle at its center. The plane passing through the centers of these two rectangles and the midpoint of their common side is perpendicular to both of rectangle planes. Therefore, this plane contains lines $l$ and $l'$. Since $l$ and $l'$ are coplanar, they meet each other at some point $O$ which has equal distance from all vertices of the polygon.
In the second case, according to the equations of the part (a), we have only one side parallel to one of axes. But this is contradiction, because we must have at least two sides parallel to each coordinate axis.

c) There exists such polygon. We will use $A_1, A_2, \dots, A_{2014}$ for its vertices. We define
$$A_n = \begin{cases} (k, k, 0) & n = 2k + 1,\ 1 \le n \le 1007 \\ (k, k - 1, 0) & n = 2k,\ 1 \le n \le 1007 \\ (503 - k, 503 - k, 1) & n = 1007 + 2k + 1,\ 1008 \le n \le 2014 \\ (503 - k, 503 - k + 1, 1) & n = 1007 + 2k,\ 1008 \le n \le 2014 \end{cases}$$
It is easy to see that the plane $2x - 2y + 2z = 1$ passes through the midpoints of all sides of the polygon.

d) Referring to the equations in the part (a), we have
$$X + Y + 2Z = 2A + 2B$$
so
$$2C + 2Z = 2A + 2B,\ C + Z = A + B$$
hence
$C < A + B$
By similar arguments, we can show $A < B + C$ and $B < C + A$. Hence, $A$, $B$ and $C$ are sides of a triangle.
For the converse, suppose that we have a triangle with sides $c \le b \le a$. We set
$n = \frac{b+c-a}{2}$. First we draw $b+a-c$ sides in the $xy$ plane parallel to the $x$ and $y$ axes, alternatively ($\frac{b+a-c}{2}$ parallel to the $x$ axis and $\frac{b+a-c}{2}$ parallel to the $y$ axis). Next, we $\frac{c+a-b}{2}$ sides parallel to $x$ and $z$ axes. Now we have $n$ number of sides parallel to the $y$ and $z$ axes remained. We can draw them to close the polygon.