Regarding question 1, any line in the Fermat cubic $C = \{X_0^3 + X_1^3 + X_2^3 + X_3^3 = 0\}$ must meet the coordinate hyperplane $H_0 = \{X_0 = 0\}$. So which points $x \in (C \cap H_0)$ can lie on lines? If $Y, Z$ are homogenous coordinates on $T_x(C \cap H_0) \cong \mathbb{P}^1$, then the restriction of $X_0^3 + X_1^3 + X_2^3 + X_3^3$ to $T_x C$ is of the form $X_0^3 + F(Y,Z)$ for a homogeneous cubic $F$. For $x$ to lie on a line, $X_0^3 + F$ must factorise, so $F$ is a cube. This means that $x$ is an inflection point of the plane cubic curve $C \cap H_0 = \{X_1^3 + X_2^3 + X_3^3 = 0\}$. The inflection points are given by intersection with the zero set of the Hessian determinant $27X_1X_2X_3$216X_1X_2X_3$. Hence , the intersection of any line in $C$ with any coordinate hyperplane must actually have two corrdinates equal to 0, and it follows that the lines consist of $\{X_0^3 + X_1^3 = X_2^3 + X_3^3 = 0\}$ and its two permutations images under permutating the coordinates (9 lines in each).
P.S. Here is a related exercise I like. Once one has identified the 27 lines in the Fermat cubic $C$, we one can use the symmetries of $C$ to guess how to arrange 6 points in $\mathbb{P}^2$ so that the blow-up is isomorphic to $C$, and then write down an explicit birational rational map $\mathbb{P}^2 \to C$dashrightarrow \mathbb{P}^3$ that realises this isomorphismmaps birationally onto $C$.
