An easy solution to the first question uses the affixes $z_k$ of the points $P_k$, since $z_0=0$ and $z_{k+1}-z_k=a_ku^{k+1}$ with $u=\mathrm e^{\mathrm i\pi/3}$. The condition that $P_6=P_0$ reads as $\sum\limits_{k=1}^6a_ku^k=0$. Introduce the imaginary part $v$ of $u$, that is,
$ v=\sqrt3/2. $ Then the real and imaginary parts of the sum above are $a_1+\frac12a_2-\frac12a_3-a_4-\frac12a_5+\frac12a_6$ and $v(a_2+a_3-a_5-a_6)$ and these are both zero if and only if:
There exists $b$ such that $a_1=a_4+b$, $a_5=a_2+b$ and $a_3=a_6+b$.
Note that this implies that $a_1+a_2+a_3+a_4+a_5+a_6=2(a_2+a_4+a_6)+3b$. From now on we reduce everything to the variables $(a_2,a_4,a_6)$ and $b$.
To solve the second and third questions, first note that opposite sides of the hexagon are parallel and imagine moving points $P_3$, $P_4$ and $P_5$ in the horizontal direction, substracting some quantity $c$ to their three abscissæ. Algebraically, this amounts to adding $c$ to $a_3$ and $a_6$ and leaving the other $a_k$s and $b$ unchanged. Geometrically, this adds $c$ times the height of the hexagon to the global area.
The height of the hexagon, in other words the distance between the lines $P_5P_6$ and $P_2P_3$, is $v(a_1+a_2)=v(a_4+a_2+b)$ hence $ A(a_2,a_4,a_6+c,b)=A(a_2,a_4,a_6,b)+cv(a_4+a_2+b), $ where $A(x,y,z,t)$ is the area of the hexagon when $a_1=y+t$, $a_2=x$, $a_3=z+t$, $a_4=y$, $a_5=x+t$, $a_6=z$ and $b=t$. In particular, $ A(a_2,a_4,a_6,b)=A(a_2,a_4,0,b)+va_6(a_4+a_2+b). $ Applying the same argument to the two other directions, one gets $ A(a_2,a_4,a_6,b)=A(0,0,0,b)+va_6(a_4+a_2+b)+va_4(a_2+b)+va_2b. $ The hexagon parametrized by $(0,0,0,b)$ is in fact the equilateral triangle with side $b$, hence $A(0,0,0,b)=vb^2/2$. This yields finally
$ A(a_2,a_4,a_6,b)=v(a_2a_4+a_4a_6+a_6a_2+b(a_2+a_4+a_6)+b^2/2). $
The second question asks for the maximal value of $A$ when $b=1$ and $a_2+a_4+a_6=\frac32$. Lagrange multipliers method indicates that this happens when $a_2=a_4=a_6$, hence the maximal value is $ A(1/2,1/2,1/2,1)=v\cdot11/4=11\sqrt3/8. $ The third question asks for the maximal value when $a_2+a_4+a_6=3s$ is such that $2s+b=2$. For each value of $s$, $a_2a_4+a_4a_6+a_6a_2$ is maximal when $a_2=a_4=a_6=s$ hence one looks for the maximal value of $3s^2+3bs+b^2/2$. Since $b=2-2s$, this is $2+s(2-s)\le 3$ which reaches its maximal value when $s=1$ and $b=0$. Finally, the maximal value corresponds to the regular hexagon with side $1$ and is $ A(1,1,1,0)=3v=3\sqrt3/2. $
Edit Using your picture, one sees that $A$ is the area of the equilateral triangle with side $a_2+a_4+a_6+2b$ minus the areas of the three equilateral triangles with side $a_1=a_4+b$, $a_3=a_6+b$ and $a_5=a_2+b$. The area of an equilateral triangle with side $\ell$ being $\sqrt3\ell^2/4$, one gets directly $ A(a_2,a_4,a_6,b)=\left((a_2+a_4+a_6+2b)^2-(a_4+b)^2-(a_2+b)^2-(a_6+b)^2\right)\sqrt3/4. $ Expanding the squares, one falls back on the formulas above.