In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$? Given dual numbers, what would be the value of $0^\varepsilon$ and $\varepsilon^\varepsilon$?
I suspect that neither one of your expressions has a canonical value.  Exponentials can be tricky because the operation of exponentiation conflates two (and a half) loosely-related concepts: the family of power functions $f_n(x):x\to x^n$ that raise $x$ to the $n$th power (and more abstractly, the notion of 'exponential object' in a category that considers $A^B$ as the collection of all the functions from $A$ to $B$) and the (canonical) exponential map $\operatorname{exp}(x):x\to e^x$ that (in a very abstract sense) maps 'local to global'.

For any ring $\mathcal{R}$, the family of functions $f_*$ has a straightforward, canonical definition as a set of functions from $\mathcal{R}$ to $\mathcal{R}$: $f_n$ is just the $n$-times multiplication of $x$ with itself.  (On the other hand, these functions don't necessarily have inverses: $f_2$ - that is, the squaring function - is a map from $\mathbb{Z}\to\mathbb{Z}$, but there's no $n\in\mathbb{Z}$ with $f_2(n)=2$.)  More broadly, this idea leads to the aforementioned notion of an exponential object in a category as a means of organizing functions from $A$ to $B$ - in this sense, $n^2$ just counts the way you can pick two (ordered, not necessarily distinct) numbers between $1$ and $n$, and likewise $2^n$ counts the number of ways you can pick $n$ 'bits'.

On the other hand, the exponential map is a fundamentally distinct concept; whereas the power maps have range a subset of their domains (since multiplication is a map $\mathcal{R}\times\mathcal{R}\to\mathcal{R}$), the range of an exponential map can be entirely distinct from its domain.  For instance, the exponentiation map on a Lie algebra takes elements of that algebra to their 'exponentiations' in a Lie group; see http://en.wikipedia.org/wiki/Exponential_map#Lie_theory for some details on this case.  While the two structures are related, they're fundamentally distinct; in a sense the Lie algebra studies local structure, while the Lie group looks at global structure, and exponentiation serves as a sort of 'integration' to show how the local structure extends out into a global structure.

A classic example of this is rotation (for simplicity, about the origin): on a 'local' (or 'instantaneous') level a rotation in three dimensions picks out an axis (the axis being rotated around) and a velocity; in two dimensions it just picks out a velocity (the speed of rotation).  The exponential map then takes this to a transformation of the underlying two-dimensional space; it serves as a map (e.g.) $\operatorname{rot}:\mathbb{R}\to SL_2(\mathbb{R})$ from real numbers to $2\times 2$ matrices of real numbers with determinant $1$ that integrates out the 'instantaneous rotation' to the resulting transformation after one (abstract) unit of time.  Note that even though this can be written as $e^X$ (and that it happily can be computed using the power series for exponential in this case), it's not a function that takes the two arguments $e$ and $X$; and raises one to the power of the other; it's fundamentally a function that takes a single argument $X$ and gives you its exponential.

The fact that these two distinct concepts of exponentiation (essentially) happen to agree over the real numbers (and that there's a real number $e$ such that $\operatorname{exp}(x) = e^x$) is, in some sense, a deep coincidence; it's not something that you can expect to happen anywhere else.  In particular, this means that you shouldn't expect expressions like $\varepsilon^\varepsilon$ to be defined on the dual numbers, because the definition of exponentiation that shows up in the duals is essentially the exponential map rather than an exponential object; its exponentiation is a one-argument function, not a two-argument one, so the expression itself doesn't make sense.
I suspect that $0^\epsilon=0$. 

If we define the exponent $\exp(x)$ in the usual way $\exp(x):= 1 + x +\frac{x^2}{2!} + \ldots$ then in the duals for $ k \cdot \epsilon$ is $\exp(k\, \epsilon) = 1 + k\,\epsilon $.
http://en.wikipedia.org/wiki/Dual_number

Then if we use $a^b= \exp (\log(a) b) $ we get $\epsilon^\epsilon= 1+ \log (\epsilon) \epsilon$.
However, I am not sure that $\log( \epsilon)$ is well defined.