Answer 1. Take the projective line $\mathbb{P}^{1}_{k}$, and let $p$ be the correspondence projecting to a point. Then this has an image, namely $\mathrm{Spec}(k)$. However, it does not have a kernel. One can see this by looking at cohomological realisations. If the kernel existed, it would only have an $\mathrm{H}^{2}$; since the $\mathrm{H}^{0}$ is accounted for by $\textrm{Spec}(k)$.
[Edit] I realise I am made this more difficult then necessary. Suppose that all idempotents have kernels. Then the category of correspondences is equivalent to its Karoubian envelope. In the Karoubian envelope we have a decomposition $\mathbb{P}^{1}_{k} = \textrm{ker}(p) \oplus \textrm{ker}(1 - p)$. The direct sum in this category is given by the disjoint union of the underlying schemes and correspondences. Under our assumption of equivalence with the category of correspondences, this exhibits the connected scheme $\mathbb{P}^{1}_{k}$ as the disjoint union of $\textrm{Spec}(k)$ and some other scheme. Contradiction. [/Edit]
Answer 2. Since the Hom-sets have $\mathbb{Q}$-coefficients, every isogeny becomes an isomorphism. But maybe you find this answer a bit cheating.