Why the words "inner" and "outer" to designate products? Does anyone know what's the rationale for using the adjectives inner and outer for certain algebraic products?
Also, I've seen the term exterior algebra.  Does the exterior here have anything to do with the outer of outer product?  If so, is there an interior algebra corresponding to inner products?
 A: Take the direct product of two groups. In the inner version, you are given a group $G$, and two subgroups $H, K$, such that $H \cap K = \{ 1 \}$, and $H, K$ commute elementwise. Then the subgroup $\langle H, K \rangle$ is called the inner direct product of $H$ and $K$, as everything takes place inside $G$.
Alternatively, if you are only given two groups $H$ and $K$, you may build a group $P$ on the set $H \times K$, with the multiplication rule $(h_1, k_1) (h_2, k_2) = (h_1 h_2, k_1 k_2)$. This is called the outer direct product, as you may say that you go out of $H$ and $K$ to build $P$. 
Now $H' = \{ (h, 1) : h \in H\}$ is a subgroup of $P$ isomorphic to $H$, and $K' = \{ (1, k) : k \in K\}$ is a subgroup of $P$ isomorphic to $K$. One checks that $H' \cap K' = \{ 1 \}$, that $H', K'$ commute elementwise, and that the inner product of $H', K'$ inside $P$ is isomorphic to the outer direct product of $H, K$.
A: I have no idea what the real reason is. But elements of exterior algebra quite often represents sub-spaces of vector space and exterior product of two elements represents union of these two sub-spaces and inner product represents their intersection. 
Hence the names: 
inner - biggest subspace inside
exterior - smallest subspace outside
A: If you think of column vectors as matrices you can't multiply them directly, because they are the wrong shapes.  You have to take the transpose of one of them.  If you take the inner product 
$$\left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)\cdot\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right) = \left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)^T\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right)$$ 
The $T$ goes on the inside, if you take the outer product
$$\left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)\otimes\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right) = \left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right)^T$$ 
the $T$ goes on the outside.
The idea has been abstracted from column vectors to other areas, but the name stuck.
A: The exterior product has a direct relation with the outer product in dimension $3$.  Consider the fundamental relations:
\begin{aligned}
i \times i &= 0\\
j \times j &= 0\\
k \times k &= 0\\
i \times j &= k\\
j \times k &= i\\
k \times i &= j
\end{aligned}
Now, look at the relations for $1$-forms:
\begin{aligned}
dx \wedge dx &= 0\\
dy \wedge dy &= 0\\
dz \wedge dz &= 0
\end{aligned}
Now, associate $dx, dy, dz$ with  $i, j, k$, and associate $dx \wedge dy, dy \wedge dz, dz \wedge dx$ also with $i, j, k$.  That's just duality.
