Deonote $p$ the determinant polyonomial. Observing that $p$ is of degree one in $x_{ij}$ for every $(i,j)$.
Now we can prove $p$ is irreducible. Suppose $p=fg$. Consider $x_{11}$. Suppose $x_{11}$ appears in $f$, then $f$ is of degree one in $x_{11}$ and $g$ is of degree zero in $x_{11}$. Now consider $x_{1j}$, then $x_{1j}$ must appear in $f$, otherwise $g$ is of degree one in $x_{1j}$ and $f$ is degree zero in $x_{1j}$, then the equality
$$fg=(ax_{11}+b)(cx_{1j}+d)=acx_{11}x_{1j}+bcx_{1j}+adx_{11}+bd\in \mathbb{F}[x_{11}, x_{1j}, \dots]$$
leads to contradiction. So all $x_{1j}$ in $f$ for $j=1,\ldots,n$. Similar $x_{j1}$ are all in $f$. And since $x_{j1}$ is in $f$, it follows $x_{jk}$ are in $f$. Finally, all $x_{ij}$ are in $f$. And $g$ is a constant. We are done!
Edit:
Contradiction: view $p$ be a polynomial of $x_{11},x_{1j}$, then $$p=x_{11}h_1+x_{1j}h_2+h_3\in \mathbb{F}[x_{11},x_{1j}, \dots],$$ where $h_1,h_2,h_3 \in \mathbb{F}[\{x_{ij}\}\mid x_{ij}\neq x_{11},x_{1j}]$, i.e., they are "constant" about $x_{11},x_{1j}$, but $$fg=acx_{11}x_{1j}+bcx_{1j}+adx_{11}+bd,$$ while $0\neq ac \in \mathbb{F}[\{x_{ij}\}\mid x_{ij}\neq x_{11},x_{1j}]$ and $bc,ad,bd$ are "constant" about $x_{11},x_{1j}$(all the results come from the assumption $f$ is a polynomial of degree one in $x_{11}$ and of degree zero in $x_{1j}$ and $g$ is of degree one in $x_{1j}$ and of degree zero in $x_{11}$), so $p$ cannot equal to $fg$ since the definition of the determinant.
Here is a non-so-explicit answer using linear algebra : it is a standard fact that the tangent space of a Grassmannian $X = G(k,V)$ at $W$ is canonically identified with $Hom(W,V/W)$, so the cotangent bundle at $W$ is $Hom(V/W,W)$.
Let us pick a supplementary $W' \oplus W = V$ and identify $V/W \cong W'$. Let $p : V = W \oplus W' \to W'$ and $q : V = W \oplus W' \to W$ be the projections. We get an isomorphism $$ Hom(V/W,W) \cong \{Z \subset V : \dim Z = k, p_{|Z} : Z \to W' \text{ is an isomorphism } \}$$
given by $f \mapsto Graph(f)$ and $ Z \mapsto (p_{Z})^{-1} \circ q$. So picking generic $Z_1, \dots, Z_{n(n-k)}$ gives you an element of $\bigwedge^{top}T_W^*X$. We take them generic because I assume you don't want the zero section :-)
Now let $$U_i = \{W \in X : p_{|Z} : Z \to W' \text{ is an isomorphism}\}$$ $$U = \{W \in X : W \oplus W' = V \}$$
Clearly on $\tilde U = U \cap U_1 \cap \dots \cap U_{k(n-k)}$ each of the $Z_i$ is a well defined section of $T^*X$ using $W'$ so $Z_1 \wedge \dots \wedge Z_{k(n-k)}$ is a local section of $\omega_X$ as expected.
Best Answer
(1) If you ask that $E$ and $H$ both be nonsingular, the answer is no. In $\mathbb{P}^3$, let $Q$ be a degree $2$ hypersurface with a singularity at one point $p$ (the cone on a smooth conic). Let $C$ be a generic smooth cubic; so $C$ does not pass through $p$ and is transverse to $Q$. Then $X:=C \cap Q$ is a degree $6$, genus $4$, smooth curve, embedded by its canonical section.
If we want to write $X$ as $A \cap B$, for hypersurfaces $A$ and $B$, then we must have $(\deg A)(\deg B) = 6$. Since $X$ does not lie in any plane, one of $A$ and $B$ must have degree $2$. There is only one degree two hypersurface containing $X$, namely $Q$, and $Q$ is not smooth.
In fact, I can give a smaller example. In $\mathbb{P}^2$, take a smooth cubic $C$ and intersect it with $Q$, the union of two lines, where the crossing point of the lines is not in $C$. You get $6$ points, and the same logic applies.
(2) If you work in $\mathbb{P}^n$, over a characteristic zero field, and only ask that $E$ be smooth (and not $H$), the answer is yes. Let $X = \{ f_1 = f_2 = \cdots = f_r =0 \}$ with $\deg f_i = d_i$, and reorder the terms such that $d_1 \geq d_2 \geq \cdots \geq d_r$. Choose $h_{ij}$ a polymonial of degree $d_i - d_j$, and otherwise generic. Define $g_i = f_i + \sum_{j>i} h_{ij} f_j$. Clearly, the $g_i$ and $f_i$ generate the same homogeneous ideal. We claim that, for generic values of the $h_{ij}$, every intersection $\{ g_1 \cap g_2 \cap \cdots \cap g_s =0 \}$ is smooth, for $s \leq r$.
Proof: By induction on $s$. When $s=0$, the claim is trivial. By induction, fix $h_{ij}$ for $i<s$ such that $\{ g_1 \cap g_2 \cap \cdots \cap g_{s-1} =0 \}$ is smooth; call this smooth variety $A$. We want to show that for generic $h_{sj}$, the intersection $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$ is smooth.
We now use a variant of Corollary III.10.9 in Hartshorne. (This is where we use the characteristic zero hypothesis.) Look at the linear system on $A$ spanned by $f_s$ and by $f_j \mathcal{O}(d_s-d_j)|_A$. The variant we want is that, over a field of characteristic zero, on a smooth variety, a generic element of a finite dimensional linear system has singularities contained within the basepoints of the linear system. Hartshorne sketches the proof of this in Remark 10.9.2. (Hartshorne fails to mention that the algebraic closure hypothesis can be removed, but I think that's just an oversight on his part.) So, for generic $h_{sj}$, the singularities of $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$ are contained in $X$.
Suppose for the sake of contradiction that $x \in X$ is a singular point of $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$. But then $X = A \cap \{ f_s + \sum h_{sj} f_j=0 \} \cap \{ f_{s+1} = f_{s+2} = \cdots = f_r =0 \}$. Slicing a singular point by a regular sequence leaves a singular point behind, so this shows that $X$ is singular at $x$, contradicting the hypothesis that $X$ is smooth. QED
(3) If you take $X=\mathbb{P}^2 \times \mathbb{P}^2$, the answer is no. Take the intersection of a $(2,1)$ hypersurface and a $(1,2)$ hypersurface, each with an isolated singularity and meeting transversely. The intersection is a smooth surface $S$. The coholomogy ring of $\mathbb{P}^2 \times \mathbb{P}^2$ is $\mathbb{Z}[u,v]/\langle u^3, v^3 \rangle$.
Suppose we could write $S$ as the intersection of a hypersurface of degree $(a_1, a_2)$ and one of degree $(b_1, b_2)$. Then, in $H^2(\mathbb{P}^2 \times \mathbb{P}^2)$, we have $(a_1 u + a_2 v)(b_1 u + b_2 v) = (2 u + v)(u + 2 v)$. By unique factorization of polynomials, we must have $(a_1, a_2) = (2,1)$ and $(b_1, b_2) = (1,2)$, or vice versa. (Note that I had to go up to $\mathbb{P}^2 \times \mathbb{P}^2$, so that the degree $2$ polynomials would inject in $H^*$; this argument doesn't work in $\mathbb{P}^1 \times \mathbb{P}^1$.)
But the only $(1,2)$ form which vanishes on $S$ is the singular $(1,2)$ form we started with, and the same for $(2,1)$ forms. So $S$ cannot be written as a complete intersection with either factor being smooth. QED
I couldn't figure out:
(4) Characteristic $p$, in $\mathbb{P}^n$. I tried to build a counter-example based on Hartshorne Exercise III.10.7 and failed. I don't have an intuition for whether that is because I was not clever enough, or because the result is true.