Number of tangent lines, inflection and singular points on a pencil of cubics $\Lambda$ 
Given the pencil of curves $\Lambda$ generated by the cubics of equations 
  $$F = 2X_0^2X_2 - 4X_0X_1^2+X_0X_1X_2+X_1^2X_2$$
  and
  $$G=4X_0^2X_2-4X_0X_1^2+X_0X_1X_2-X_1^2X_2$$
  find for which of the cubics in $\Lambda$, the coordinate points $(1:0:0),(0:1:0),(0:0:1)$ are singular. Show if one of the coordinate points is an inflection point for some cubic in $\Lambda$. Show that $\Lambda$ can be generated by reducible cubics. Find how many tangents to $V(F)$ go through the point $(1:-4:0)$.

For the first question, I just computed the partial derivatives of $F$ and $G$ and looked for which coordinate points, all the partial derivatives are zero. I found that $(1:0:0)$ is a singular point of multiplicity $2$ for both $V(F)$ and $V(G)$, but I'm not sure if that immediately proves that there isn't any other cubic in $\Lambda$ such that some other point isn't singular.
For the second question, I computed the Hessian of $f$ and $G$ on the coordinate points that are not singular and I found that $(0:1:0)$ is an inflection point for $V(F)$ and $(1:0:0)$ is an inflection point for $V(G)$, but again, I'm not sure if that answers the question for all the cubics in $\Lambda$.
Now, the question about proving that $\Lambda$ can be generated by reducible cubics, I simply have no idea about how to proceed.
For the last, about the number of tangent lines to $V(F)$ that goes through the point $(1:-4:0)$, I remembered I read somewhere that I had to find the polar curve to that point and the intersecting with the curve, so that means finding the points that satisfy
$$F_0(X_0, X_1, X_2) - 4F_1(X_0,X_1,X_2) = 0$$
and
$$F(X_0,X_1,X_2)=0$$
but I don't know if this is correct, or if there's another way to find the tangent lines to a curve $C$ that go through a point $P$. Thank you in advance for your help!
P.D. I'm assuming that the line at infinity on $\mathbb{P}_2$ is $X_0 = 0$ instead of $X_2 = 0$.
 A: The pencil is $\Lambda = \{ C_\lambda | C_\lambda = \{ \lambda_0 F + \lambda_1 G = 0\}, \lambda = (\lambda_0: \lambda_1) \in \mathbb P^1 \}$.
To the first part:
One has to take $C_\lambda \in \Lambda$ and compute the gradient of the defining function. So,
$$
\nabla_{\lambda_0 F + \lambda_1 G} = \begin{pmatrix}\lambda_0(4X_0X_2 - 4X_1^2+X_1X_2) + \lambda_1(8X_0X_2-4X_1^2 + X_1X_2) \\
\lambda_0(-8X_0X_1 + X_0X_2 + 2X_1X_2) + \lambda_1 (-8X_0X_1 + X_0X_2 - 2X_1X_2) \\
\lambda_0(2X_0^2 + X_0X_1 + X_1^2) + \lambda_1 (4X_0^2 + X_0X_1 - X_1^2)
\end{pmatrix}.$$
Plugging in the three coordinate points, one gets the following conditions:
I) $\lambda_0 + 2\lambda_1 = 0$ by plugging in $(1:0:0)$ (note that the third component of the gradient does not vanish after plugging in this point).
II) $\lambda_0 + \lambda_1 = 0$ and $\lambda_0 - \lambda_1 = 0$ by plugging in $(0:1:0)$ (looking at first and third component of gradient).
III) Empty conditions, as $(0:0:1)$ is a singular point of both $V(F)$ and $V(G)$.
Here you can read off for which $\lambda$ the curve $C_\lambda$ has the desired properties:
I) $(1:0:0)$ is singular if and only if $\lambda = (-2:1)$.
II) $(0:1:0)$ is never singular.
III) $(0:0:1)$ is always singular.
To the "inflection point" part: 
Just compute it in the same manner as I did in the first part. Compute the Hessian of $C_\lambda$, and determine the values of $\lambda$ for which the respective point lies on the Hessian and are smooth. I leave this to you.
$\Lambda$ can be generated by reducible cubics:
Note that a pencil of cubics is just a $1$-dimensional projective subspace of the projective space $\mathbb P (V)$, where $V$ is the vector space of all homogeneous polynomials of degree $3$. Whence we can write $\Lambda = \mathbb P(U)$, where $U$ is a $2$-dimensional linear subspace of $V$. In our particular case, we have the basis $F,G$ of $U$. The exercise therefore is just to find a different basis $F_1, G_1$ of $U$, such that $V(F_1), V(G_1)$ are reducible.
For instance, I can set $$F_1 := F-G = -2X_0^2 X_2 + 2X_1^2 X_2,$$ which obviously defines a reducible curve (as the defining polynomial is divisible by $X_2$). Setting $$ G_1 := 2F-G,$$ I get another reducible curve (check!) and $F_1, G_1$ are linearly independent (check!), so form a basis of $U$.
Tangent lines of $V(F)$ through $(1:-4:0)$:
What you wrote is correct. Recall the definition of the tangent line through a smooth point $p \in V(F)$:
$$T_p = \{ \sum_{j=0}^2 \frac{\partial F}{\partial X_j}(p) \cdot X_j = 0 \}.$$
Recall as well the definition of the polar of $V(F)$ with respect to the point $q = (q_0:q_1:q_2)$:
$$H_q = \{ p \in \mathbb P^2 | \sum_{j=0}^2 \frac{\partial F}{\partial X_j}(p) \cdot q_j = 0 \}.$$
So we see that a smooth point $p \in V(F)$ is contained in $H_q$ if and only if $q \in T_p$. Geometricially speaking, computing the set $H_q \cap V(F)$ is just determining the set of points $p \in V(F)$ that are either singular or the tangent line through $p$ contains $q$.
Another hint is that it is only asked for the number of the tangent lines, not for the points in $V(F)$ the tangent lines pass. So one can use case-by-case analysis (e.g. the cases $X_1 = 0$, $X_1 = 1$) and abstract arguments, for instance "a univariate polynomial of positive degree over $\mathbb C$ has only simple zeros if and only if its discriminant is non-zero if and only if it has no common zero with its derivative".
Now you can try calculating. :-)
