You can definitely approach this from the perspective of exterior algebra. Indeed, the part of your question relating to your first identity has been asked before, so for the coordinate-free, exterior algebra proof of that, please take a look at my answer to the earlier question. So, let me know turn to your second identity. Before continuing, let me recycle some background and notation from my old answer.
Recall that if $V$ is an $n$-dimensional inner product space, then the
Hodge star is a linear isomorphism $\ast : \bigwedge^k V \to \bigwedge^{n-k} V$ for each $0 \leq k \leq n$, satisfying the
following:
for $v$, $w \in V$, $\langle v,w\rangle \omega = v \wedge \ast w$ for $\omega = \ast 1$ the generator of $\bigwedge^n V$ satisfying
$\omega = e_1 \wedge \cdots \wedge e_n$ for any orthonormal basis
$\{e_k\}$ of $V$ with the appropriate orientation (e.g., the volume
form in $\bigwedge^n (\mathbb{R}^n)^\ast$);
in particular, $\ast \ast = \operatorname{Id}$ when $n$ is odd.
Also, recall that the inner product on $\bigwedge^k V$ is given by $$\langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k \rangle = \det(\langle v_i,w_j \rangle). $$
So, suppose that $V$ is $3$-dimensional, in which case we can define
the cross product of $a$, $b \in V$ by $$a \times b := \ast (a\wedge b).$$
Now, let $a$, $b$, $c$, $d \in \mathbb{R}^3$. Then
$$
\langle a \times b, c \times d \rangle \omega = \langle \ast (a \wedge b), \ast (c \wedge d) \rangle \omega \\
= \ast (a \wedge b) \wedge \ast \ast (c \wedge d) \\
= \ast (a \wedge b) \wedge (c \wedge d)\\
= (-1)^{1 \cdot 2} (c \wedge d) \wedge \ast (a \wedge b)\\
= \langle c \wedge d, a \wedge b \rangle\omega \\
= \langle a \wedge b, c \wedge d \rangle\omega\\
= \begin{vmatrix} \langle a,c \rangle & \langle a,d \rangle \\ \langle b,c \rangle & \langle b,d \rangle \end{vmatrix}\omega\\
= (\langle a,c \rangle \langle b,d \rangle - \langle a,d \rangle \langle b,c \rangle)\omega,
$$
as was required. Observe, in particular, that the right-hand side of your second identity looks like a $2 \times 2$ determinant because once you've unpacked all relevant definitions, it really is a $2 \times 2$ determinant, namely the inner product
$$
\langle a \wedge b, c \wedge d \rangle := \begin{vmatrix} \langle a,c \rangle & \langle a,d \rangle \\ \langle b,c \rangle & \langle b,d \rangle \end{vmatrix}
$$
of the bivectors $a \wedge b$, $c \wedge d \in \bigwedge^2 \mathbb{R}^3$.
