No. Note that $V$ has dimension $4$. The maximum dimension of an elliptic subspace of $\Lambda^2(V^\ast)$ is $3$, so if $\Pi_1$ and $\Pi_2$ don't intersect, then $\Pi_1\oplus \Pi_2$ is never elliptic.

To see why this is so, let $\Omega\in \Lambda^4(V^\ast)$ be a volume form and define a symmetric bilinear form $q$ on $\Lambda^2(V^\ast)$ by the rule $$ q(\alpha,\beta)\ \Omega = \alpha\wedge\beta. $$ If $e^i$ ($0\le i\le 3$) is a basis of $V^\ast$ such that $\Omega = e^0\wedge e^1\wedge e^2\wedge e^3$, then setting $$ \alpha^1 = e^0\wedge e^1,\quad \alpha^2 = e^0\wedge e^2,\quad \alpha^3 = e^0\wedge e^3 $$ and $$ \alpha^4 = e^2\wedge e^3,\quad \alpha^5 = e^3\wedge e^1,\quad \alpha^6 = e^1\wedge e^2, $$ one gets that the matrix $Q = (Q^{ij}) = \bigl(Q(\alpha^i,\alpha^j)\bigr)$ of $q$ with respect to this basis has the block form $$ Q = \begin{pmatrix} 0_3 & I_3\\\\ I_3 & 0_3 \end{pmatrix}. $$ Clearly, $Q$ has type $(3,3)$, so $q$ cannot be definite (positive or negative) on any subspace of $\Lambda^2(V^\ast)$ that has dimension greater than $3$.

By the way, $\mathrm{SL}(\Omega)\subset \mathrm{GL}(V)$ acts transitively on the set of positive (or negative) definite subspaces of a given dimension.

No. Note that $V$ has dimension $4$. The maximum dimension of an elliptic subspace of $\Lambda^2(V^\ast)$ is $3$, so if $\Pi_1$ and $\Pi_2$ don't intersect, then $\Pi_1\oplus \Pi_2$ is never elliptic.

To see why this is so, let $\Omega\in \Lambda^4(V^\ast)$ be a volume form and define a symmetric bilinear form $q$ on $\Lambda^2(V^\ast)$ by the rule $$ q(\alpha,\beta)\ \Omega = \alpha\wedge\beta. $$ If $e^i$ ($0\le i\le 3$) is a basis of $V^\ast$ such that $\Omega = e^0\wedge e^1\wedge e^2\wedge e^3$, then setting $$ \alpha^1 = e^0\wedge e^1,\quad \alpha^2 = e^0\wedge e^2,\quad \alpha^3 = e^0\wedge e^3 $$ and $$ \alpha^4 = e^2\wedge e^3,\quad \alpha^5 = e^3\wedge e^1,\quad \alpha^6 = e^1\wedge e^2, $$ one gets that the matrix $Q = (Q^{ij}) = \bigl(Q(\alpha^i,\alpha^j)\bigr)$ of $q$ with respect to this basis has the block form $$ Q = \begin{pmatrix} 0_3 & I_3\\\\ I_3 & 0_3 \end{pmatrix}. $$ Clearly, $Q$ has type $(3,3)$, so $q$ cannot be definite (positive or negative) on any subspace of $\Lambda^2(V^\ast)$ that has dimension greater than $3$.

By the way, $\mathrm{SL}(\Omega)\subset \mathrm{GL}(V)$ acts transitively on the set of positive (or negative) definite subspaces of $\Lambda^2(V\ast)$ of a given dimension. In fact, $\mathrm{SL}(\Omega)$ acts transitively on the subspaces of a given $q$-type with one exception: There are two $\mathrm{SL}(\Omega)$ orbits in the set of $q$-null $3$-planes: The first type is the orbit of the span of $\alpha^1,\alpha^2,\alpha^3$ as described in the basis above, i.e., the set of multiples of a single $1$-form. The second type is the orbit of the span of $\alpha^4,\alpha^5,\alpha^6$ as described in the basis above, i.e., spaces of the form $\Lambda^2(W)\subset\Lambda^2(V^\ast)$ where $W$ is a hyperplane in $V^*$.

Finally, note that the connected $15$-dimensional group $\mathrm{SL}(\Omega)$ acts on $\Lambda^2(V^\ast)$ preserving $q$ and an orientation of $\Lambda^2(V^\ast)$, which induces a homomorphism $\mathrm{SL}(\Omega)\to\mathrm{SO}(q)$. This homomorphism has $\{\pm I\}\subset \mathrm{SL}(\Omega)$ as kernel and maps onto the identity component of $\mathrm{SO}(q)$ (which, itself, has two components). This is one of the 'exceptional' isomorphisms of the classical groups.

No. Take $V$ to have dimension $4$. The maximum dimension of an elliptic subspace of $\Lambda^2(V^\ast)$ is $3$, so if $\Pi_1$ and $\Pi_2$ don't intersect, then $\Pi_1\oplus \Pi_2$ is never elliptic.

No. Note that $V$ has dimension $4$. The maximum dimension of an elliptic subspace of $\Lambda^2(V^\ast)$ is $3$, so if $\Pi_1$ and $\Pi_2$ don't intersect, then $\Pi_1\oplus \Pi_2$ is never elliptic.

To see why this is so, let $\Omega\in \Lambda^4(V^\ast)$ be a volume form and define a symmetric bilinear form $q$ on $\Lambda^2(V^\ast)$ by the rule $$ q(\alpha,\beta)\ \Omega = \alpha\wedge\beta. $$ If $e^i$ ($0\le i\le 3$) is a basis of $V^\ast$ such that $\Omega = e^0\wedge e^1\wedge e^2\wedge e^3$, then setting $$ \alpha^1 = e^0\wedge e^1,\quad \alpha^2 = e^0\wedge e^2,\quad \alpha^3 = e^0\wedge e^3 $$ and $$ \alpha^4 = e^2\wedge e^3,\quad \alpha^5 = e^3\wedge e^1,\quad \alpha^6 = e^1\wedge e^2, $$ one gets that the matrix $Q = (Q^{ij}) = \bigl(Q(\alpha^i,\alpha^j)\bigr)$ of $q$ with respect to this basis has the block form $$ Q = \begin{pmatrix} 0_3 & I_3\\\\ I_3 & 0_3 \end{pmatrix}. $$ Clearly, $Q$ has type $(3,3)$, so $q$ cannot be definite (positive or negative) on any subspace of $\Lambda^2(V^\ast)$ that has dimension greater than $3$.

By the way, $\mathrm{SL}(\Omega)\subset \mathrm{GL}(V)$ acts transitively on the set of positive (or negative) definite subspaces of a given dimension.