This is the exercise III 10.7 in Hartshorne's Algebraic Geometry
I am not sure if I misunderstood the question.
The seven points of the projective plane over $\mathbb{F}_2$, I think, means
$\{[x_0,x_1,x_2]:x_i\in\mathbb{F}_2,\forall i\}$, i.e. closed points of $\mathbb{P}^2_{\mathbb{F}_2}$
I think that $X$ denotes $\mathbb{P}^2_k$
The linear system $\delta$ of all cubic curves in $X$ passing through all $P_i$, I think,
means $\delta$ corresponding to $V\subset\Gamma(X,\mathcal{O}(3))$, defined by $V=\{s\in\Gamma(X,\mathcal{O}(3)):s(P_i)=0,\forall i\}$
If I'm not misunderstanding something, then $V$ can be written down easily:
$V$ is the $k$-vector subspace with basis $x_0x_1(x_0+x_1),x_1x_2(x_1+x_2),x_2x_0(x_2+x_0)$
Under the above setting, I have worked out (a), and the second half of (b)
I'm confused about the first half part of (b)
Every $C\in\delta$ can be written as $(s=0)\subset X$
where $s=c_0x_1x_2(x_1+x_2)+c_1x_2x_0(x_2+x_0)+c_2x_0x_1(x_0+x_1)$, for some $[c_0,c_1,c_2]\in\mathbb{P}^2_k$
If $c_0=0$, $c_1,c_2\neq0$ and $c_1\neq c_2$
Then $s=x_0(c_1x_2(x_0+x_2)+c_2x_1(x_0+x_1))$
$C=(x_0=0)\cup(c_1x_2(x_0+x_2)+c_2x_1(x_0+x_1)=0)$
This quadratic cannot be written as a product of two lines. (Otherwise $\frac{c_2}{c_1}$ must be a cubic root of unity)
Then $C$ doesn't belong to the two cases mentioned in (b)
What's wrong? Had I misunderstood something?
In addition, I'd appreciate it if somebody could provide some hints on how to show that for general $[c_0,c_1,c_2]$, $C$ is irreducible cuspidal curve. The irreducible part seems not to be hard. To show it's cuspidal at the unique singular point, I need to calculate the completion of the local ring. But the equation seems to be too complicated to handle.
Everything you've written here is correct - Hartshorne's classification of the members of this linear system is incomplete. In addition to the cuspidal cubics and unions of lines, there can be a smooth conic with a tangent line as a member of this linear system, and you've demonstrated one with $V(x_0(c_1x_2(x_0+x_2)+c_2x_1(x_0+x_1)))$.
To see how to deal with part (b), note that as we're in an algebraically closed field of characteristic 2, we can write any element uniquely as a square. So we can write a curve $C$ in $\mathfrak{d}$ as $\alpha^2(x_0x_1(x_0+x_1))+\beta^2(x_0x_2(x_0+x_2))+\gamma^2(x_1x_2(x_1+x_2))$ for $\alpha,\beta,\gamma\in \overline{\Bbb F_2}$ so that the Jacobian is $$\begin{pmatrix} \alpha^2 x_1^2 + \beta^2 x_2^2 & \alpha^2 x_0^2 +\gamma^2 x_2^2 & \beta^2 x_0^2 + \gamma^2 x_1^2\end{pmatrix}$$ which means that a point $[x_0:x_1:x_2]$ on $C$ is singular iff it satisfies $\alpha x_1+\beta x_2$, $\alpha x_0 + \gamma x_2$, and $\beta x_0 + \gamma x_1$, which can be rephrased as $[x_0:x_1:x_2]$ being in the kernel of $\begin{pmatrix} 0 & \alpha & \beta \\ \alpha & 0 & \gamma \\ \beta & \gamma & 0 \end{pmatrix}$. Assuming at least one of $\alpha,\beta,\gamma$ is nonzero, this matrix is of rank 2, so there's a unique point in $\Bbb P^2$ which is a zero of all three partial derivatives, it turns out to be $[\gamma:\beta:\alpha]$ which also satisfies our equation! So there is a 1-1 correspondence between points of $X$ and members of the linear system.
To see the statement about being cuspidal, you don't actually have to compute a completion - you can compute the local equation of the curve in a patch containing the singular point and show it's of the form $y^2=x^3$. You will need some data about when/why the curve is irreducible (this depends on some data about $\{\alpha,\beta,\gamma\}$) and a few coordinate transforms (you may potentially find it helpful to look up how one takes an arbitrary elliptic curve to Weierstrass form).