Here are two ways to derive the formula for the dot product. I assume that $v_1$ and $v_2$ are vectors with spherical coordinates $(r_1, \varphi_1, \theta_1)$ and $(r_2, \varphi_2, \theta_2)$.
First way: Let us convert these spherical coordinates to Cartesian ones. For the first point we get Cartesian coordinates $(x_1, y_1, z_1)$ like this:
$$
\begin{array}{rcl}
x_1 & = & r_1 \sin \varphi_1 \cos \theta_1, \\
y_1 & = & r_1 \sin \varphi_1 \sin \theta_1, \\
z_1 & = & r_1 \cos \varphi_1.
\end{array}
$$
Similar formulas hold for $(x_2, y_2, z_2)$. Now, the dot product is simply equal to
$$
(v_1, v_2) = x_1 x_2 + y_1 y_2 + z_1 z_2 = \\
= r_1 r_2 ( \sin \varphi_1 \sin \varphi_2 ( \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) + \cos \varphi_1 \cos \varphi_2) = \\
= r_1 r_2 ( \sin \varphi_1 \sin \varphi_2 \cos (\theta_1 - \theta_2) + \cos \varphi_1 \cos \varphi_2)
$$
Second way: Actually, we could have done it without coordinate conversions at all. Indeed, we know that $(v_1, v_2) = r_1 r_2 \cos \alpha$, where $\alpha$ is the angle between $v_1$ and $v_2$. But $\cos \alpha$ can be immediately found by the Spherical law of cosines, which yields exactly the same formula that we just proved. Basically, our first way is itself a proof for the spherical law of cosines.
PS: I'm not saying anything about cross products, but my guess is that the correct formula will look terrible. Not only will it contain sines and cosines, it is likely that it will also contain arc functions (they will appear when we try to convert the result back to spherical coordinates). Unless those arc functions magically cancel out with all the sines and cosines. But it is highly unlikely, and I don't feel like going through the trouble of checking.
PPS: One more thing. Cross products are not the only scary thing about spherical coordinates. If you think about it, even addition of two vectors is extremely unpleasant in spherical coordinates. Multiplication by a number is alright though, because it only changes $r$ and doesn't affect $\varphi$ and $\theta$ (at least when we multiply by a positive number).
They are not equal: $\vec a\wedge\vec b$ is a 2-vector, while $\vec a\times \vec b$ is just a vector. They are related by the Hodge dual operator:
$$
\star: \vec e_i\wedge\vec e_j\mapsto \text{sgn}(\sigma)\vec e_k
$$
where $\sigma$ is the permutation $(1,2,3)\mapsto(i,j,k)$.
In Clifford algebra $\mathcal{Cl}_3$, they are related by:
$$
\vec a\wedge\vec b=(\vec a\times\vec b)\vec e_{123},\quad \vec e_{123}=\vec e_1\vec e_2\vec e_3.
$$
Here, the Clifford product is defined by:
$$
\vec e_i\vec e_j=\begin{cases}
-\vec e_j\vec e_i & i\neq j\\
1 & i=j
\end{cases}
$$
Knowledge of such can be easily picked up from Lounesto's Clifford Algebras and Spinors.
Best Answer
The radius vector $\vec{r}$ in cylindrical coordinates is $\vec{r}=\rho\hat{\rho}+z\hat{z}$. Calculating the cross-product is then just a matter of vector algebra:
$$\vec{a}\times\vec{r} = a\hat{z}\times(\rho\hat{\rho}+z\hat{z})\\ =a(\rho(\hat{z}\times\hat{\rho})+z(\hat{z}\times\hat{z}))\\ =a\rho(\hat{z}\times\hat{\rho})\\ =a\rho\hat{\phi},$$
where in the last line we've used the orthonormality of the triad $\{\hat{\rho},\hat{\phi},\hat{z}\}$.