Fixed a primitive root $g$ of $p$. Choose $A=\{ g^{3i-2}\mid i\in \{ 1,2,...,\frac{p-1}{3}\} \} ,B=\{ g^{3i-1}\mid i\in \{ 1,2,...,\frac{p-1}{3}\} \} $ and $C=\{ g^{3i}\mid i\in \{ 1,2,...,\frac{p-1}{3}\} \} $. Since, from a), $L(g^i)=L(g^{i+3j})$ for any $i,j\in \mathbb{Z}^+$, we can easily deduce that this partition works.

Recall that $L(m)=\sum_{i=0}^{p-1}{\Big( \frac{i(i^3+m)}{p}\Big) } =\sum_{i=1}^{p-1}{ \Big( \frac{i(i^3+m)}{p}\Big) } $. From b), we've $\sum_{m=1}^{p-1}{L(m)}=\frac{p-1}{3}(a+b+c)$. On the other hand, \begin{align*} \sum_{m=1}^{p-1}{L(m)} &=\sum_{m=1}^{p-1}{\sum_{i=1}^{p-1}{ \Big( \frac{i(i^3+m)}{p}\Big) } }=\sum_{i=1}^{p-1}{\sum_{m=1}^{p-1}{ \Big( \frac{i}{p}\Big) \Big( \frac{i^3+m}{p}\Big) } }\\ & =\sum_{i=1}^{p-1}{\Big( \frac{i}{p}\Big) \Big( \sum_{j=0}^{p-1}{\Big( \frac{j}{p}\Big) } -\Big( \frac{i^3}{p}\Big) \Big) }\\ & =\sum_{i=1}^{p-1}{ -\Big( \frac{i}{p}\Big) \Big( \frac{i^3}{p}\Big) }\\ & =\sum_{i=1}^{p-1}{-1}=p-1. \end{align*}Hence, $a+b+c=-3$.

From b), we've $\sum_{m=0}^{p-1}{L(m)^2}=(p-1)^2+\frac{p-1}{3}(a^2+b^2+c^2)$. On the other hand, \begin{align*} \sum_{m=0}^{p-1}{L(m)^2} &=\sum_{m=0}^{p-1}{\Big( \sum_{i=0}^{p-1}{ \Big( \frac{i(i^3+m)}{p}\Big) } \Big)^2 } =\sum_{m=0}^{p-1}{\sum_{0\leq i,j\leq p-1}{ \Big( \frac{i}{p}\Big) \Big( \frac{j}{p}\Big) \Big( \frac{i^3+m}{p}\Big) \Big( \frac{j^3+m}{p}\Big) }}\\ & =\sum_{0\leq i,j\leq p-1}{\sum_{m=0}^{p-1}{ \Big( \frac{i}{p}\Big) \Big( \frac{j}{p}\Big) \Big( \frac{i^3+m}{p}\Big) \Big( \frac{j^3+m}{p}\Big) }}\\ & =\sum_{0\leq i,j\leq p-1}{ \Big( \frac{i}{p}\Big) \Big( \frac{j}{p}\Big) h_{i,j} } \end{align*}where $h_{i,j}= \begin{cases} -1, & \text{if } p\nmid (i^3+j^3)^2-4i^3j^3 \\ p-1, & \text{if } p\mid (i^3+j^3)^2-4i^3j^3 \end{cases}$. Note that, for $1\leq i,j\leq p-1$, $p\mid (i^3+j^3)^2-4i^3j^3 \iff p\mid i^2+ij+j^2$ or $i=j$. Note that we used the fact that, for any $a,b,c\in \mathbb{Z}$ that $p\nmid a$, $$\sum_{x=0}^{p-1}{\Big( \frac{ax^2+bx+c}{p}\Big) } = \begin{cases} -( \frac{a}{p}) , & \text{if } p\nmid b^2-4ac \\ (p-1)( \frac{a}{p}) , & \text{if } p\mid b^2-4ac \end{cases}.$$Since $\Big( \frac{-3}{p}\Big) =1$ (not hard to prove by law of quadratic reciprocity), in $\mathbb{Z}_p$, the equation $x^2+x+1=0$, which is equivalent to $(2x+1)^2=-3$, has two solutions, called $z_1$ and $z_2$. We've $z_1+z_2=-1$ and $z_1z_2=1$. Not hard to show that $z_1\neq z_2$ and $\{ z_1,z_2\} =\{ \frac{-1+\ell}{2} ,\frac{-1-\ell}{2}\}$ where $\ell \in \mathbb{Z}_p$ that $\ell^2 =-3$. It's not hard to verify that $\frac{-1+\ell}{2} =\Big( \frac{p+1}{2}\ell +\frac{p+1}{2}\Big)^2$. So, $z_1$, and so $z_2$, is quadratic residue modulo $p$. Also, we get that, for any $1\leq i,j\leq p-1$ that $p\mid i^2+ij+j^2$, $j=z_1i$ or $j=z_2i$. And lastly, there not exists $1\leq i,j\leq p-1$ that $p\mid i^2+ij+j^2$ and $i=j$ simultaneously. Hence, continue our summation \begin{align*} \sum_{0\leq i,j\leq p-1}{ \Big( \frac{i}{p}\Big) \Big( \frac{j}{p}\Big) h_{i,j}}& =-\sum_{1\leq i,j\leq p-1}{\Big( \frac{i}{p}\Big) \Big( \frac{j}{p}\Big) }+p\Big( \sum_{1\leq i,j\leq p-1,p\mid i^2+ij+j^2}{ \Big( \frac{i}{p}\Big) \Big( \frac{j}{p}\Big) }+\sum_{1\leq i,j\leq p-1,i=j}{ \Big( \frac{i}{p}\Big) \Big( \frac{j}{p}\Big) }\Big) \\ & = -\Big( \sum_{i=1}^{p-1}{ \Big( \frac{i}{p}\Big) }\Big)^2+p\Big( \sum_{i=1}^{p-1}{\Big( \frac{i}{p}\Big)\Big( \Big( \frac{iz_1}{p}\Big) +\Big( \frac{iz_2}{p}\Big) \Big) } +(p-1)\Big) \\ & = -0+p\Big( \sum_{i=1}^{p-1}{(1+1)}+(p-1)\Big) =3p(p-1). \end{align*}Hence, $(p-1)^2+\frac{p-1}{3}(a^2+b^2+c^2)= 3p(p-1)\implies a^2+b^2+c^2=6p+3$.

From c) and d), we easily get that $a\equiv b\equiv c\pmod{3}$, this gives $X,Y\in \mathbb{Z}$. And $X^2+XY+Y^2=\frac{a^2+ab+b^2}{3}+a+b+1=\frac{(-3-c)^2+(6p+3-c^2)}{6}+(-3-c)+1=p$, done.