Several things before I answer.
a) You should really put more effort into these questions. Put them as four separate questions and show your thinking for all of them.
b) What course notes are these from? I'm just curious.
(1) As the note writer notes, $E$ is irreducible since
$$f(x,y):=y^2+a_1xy+a_3y-(x^3+a_2x^2+a_4x+a_6)$$
is irreducible. For simplicity let's assume the characteristic $k$ is different than $2$. To see this note that if we write
$$f(x,y)=g(x,y)h(x,y)$$
that $g$ and $h$ must be monic (up to scalars in $k$) as polynomials in $y$ since $f$ is. This then implies that each $g(x,y)$ and $h(x,y)$ if not constant, have degree at least $1$ in $y$. This then implies that $g(x,y)$ and $h(x,y)$ are degree $1$ in $y$. But this is absurd since that implies that
$$\frac{-(a_1x+a_3)\pm \sqrt{(a_1x+a_3)^2+4(-a_3+x^3+a_2x^2+a_4x+a_6)}}{2}$$
is a polynomial in $x$, which is clearly impossible considering that the argument of the square root is a monic odd-degree polynomial.
Now, since $f$ is irreducible we know that $V(f)\subseteq \mathbb{A}^2_k$ is irreducible. Since $E$ is the closure of $V(f)$ in $\mathbb{P}^2_k$, and closure preserves irreducibility, we deduce that $E$ is irreducible.
(2) Let $F$ denote the homogenization of $f$. So,
$$F(x,y,z)=y^2z+a_1 xyz+a_3yz^2-(x^3+a_2x^2z+a_4xz^2+a_6z^3)$$
so then $E=V(F)\subseteq\mathbb{P}^2_k$. We then know from the Jacobian criterion that $E$ is smooth iff
$$F_x=F_y=F_z=F=0$$
has no common solution in $\overline{k}$. Note though that $0$ of $E$ is the point $[0:1:0]$ and plugging this into $F_z$ results in $1$. So, $0=[0:1:0]$ can never be a singular point. Thus, it suffices to check the smoothness of $E-\{0\}$ which is the affine curve $V(f)\subseteq\mathbb{A}^2_k$.
(3) I think the note writer means 'Bezout's formula' which says that if $C$ is a geometrically integral smooth curve in $\mathbb{P}^2_k$ of degree $d$ then
$$g(C)=\frac{(d-1)(d-2)}{2}$$
This formula, as the quoted sentence suggests, comes the from the classifiation of line bundles on $\mathbb{P}^2_k$ and a cohomology calculation. In particular, if $d=3$ we get that $g(C)=1$. So, in our case $E$ has degree $3$ so that $E$ has genus $1$, thus $(E,0)$ is an elliptic curve.
EDIT: Oh, the note taker is claiming to that Bezout's formula as I said above follows from Bezout's theorem. I understand. My suggested method above calculates the arithmetic genus of $C$ (which is the same as the geometric genus by Serre duality). Namely, the adjunction formula says that
$$\omega_C=i^\ast(i_\ast\mathcal{O}_C\otimes \omega_{\mathbb{P}^2_k})$$
where $i$ is the inclusion of $C$ into $\mathbb{P}^2_k$. So then, one sees that using the degree of the canonical bundle is $2g-2$ and that $\omega_{\mathbb{P}^2_k}=\mathcal{O}_{\mathbb{P}^2_k}(-3)$ that
$$2g-2=\deg(\omega_C)=\deg(i^\ast(i^\ast(i_\ast\mathcal{O}_C\otimes \omega_{\mathbb{P}^2_k})))=\deg(C .(C-3))$$
But, if $C$ is cut out be a degree $d$ curve then $\deg(C)=d$ and so applying Bezout's theorem to the above we get
$$2g-2=d(d-3)$$
solving for $g$ gives
$$g=\frac{d(d-3)}{2}+1$$
(4) Your sections are $(x,y,1)$. The map $E\to\mathbb{P}^2_k$ can then imprecisely be written as
$$E \ni e\mapsto [x(e):y(e):1(e)]$$
where even though $x,y,1$ are only sections of a line bundle, they make sense since scalar multiplication doesn't affect points in $\mathbb{P}^2_k$ and thus it doesn't matter what chart you compute this on.
Anyways, $x(e)$ and $y(e)$ have poles of order $2$ and $3$ respetively at $0$ and $1$ has no pole at $0$. So, to evaluate $[x(0):y(0):1(0)]$ you need to multiply through by a uniformizer cubed. Let's call this uniformizer $\pi$. So then, really what $[x(0):y(0):1(0]$ means is something like $[\pi^3 x(0),\pi^3 y(0):\pi^3 1(0)]$ where now since $\pi^3x, \pi^3y$ and $\pi^3 1$ no longer have poles at $0$ it makes sense to evaluate them there. But, note that $\pi^3x$ and $\pi^3 1$ now have poles of order $-1$ and $-3$ at $0$ or, in other words, zeros at $0$. So, $\pi^3x(0)=\pi^31(0)=0$. Since $y$ had a pole of order $3$ we see that $\pi^3y$ is non-vanishing at $0$. So $[x(0):y(0):1(0)]$ becomes something like $[0:c:0]$ where $c$ is non-zero. This is just $[0:1:0]$.
