Just take a problem whose Turing degree is above $0'$, which is the degree of The Halting Oracle. In terms of the arithmetical hierarchy you want problems which are above $\Sigma^0_1$. Examples of such problems (where $\phi_n$ is the $n$-th partial computable function and $W_n = \{k \in \mathbb{N} \mid \text{$\phi_n(k)$ is defined}\}$ is the $n$-th computably enumerable set):
- $\{n \in \mathbb{N} \mid \text{$\varphi_n$ terminates for finitely many inputs}\}$ is $\Sigma^0_2$-complete.
- $\{n \in \mathbb{N} \mid \text{$\varphi_n$ is a total function}\}$ is $\Pi^0_2$-complete.
- $\{n \in \mathbb{N} \mid \text{$W_n$ is a computable set}\}$ is $\Sigma^0_3$-complete.
None of these can be solved even if you have a Halting Oracle. For instance, consider the second example, "is $\varphi_n$ total?" Given $n$ how would the Halting Oracle help us decide whether the Turing machine encoded by $n$ halts on every input?
[Added 2014-06-03] For a "practical" aspect of all of this, consider the problem: a programmer has written a function void charge_credit_card(int card_number, int amount) and we would like to know whether the function terminates on all inputs. It is impossible to write a compiler which can automatically check this in general. Moreover, even if we allow the compiler to ask us questions of the form "does charge_credit_card terminate when given input (k,m)?", it is still impossible.
