Here's an answer for 1: given a LCA group $T$, $\text{Hom}(G,T)$ is LC for every LCA $G$ iff $T$ is a compact Lie group.

Proof: Clearly this is a sufficient condition, because it holds for connected tori, and is obviously stable under taking closed subgroup.

Conversely suppose the condition is satisfied.
Let $\mathbf{Z}^{(\mathbf{N})}$ denote the discrete free abelian group of countable rank. Then $\text{Hom}(\mathbf{Z}^{(\mathbf{N})},T)=T^{\mathbf{N}}$ is locally compact iff $T$ is compact.

So $T$ is compact. Let now $D$ be the Pontryagin dual of $T$, so that $D$ is discrete. By the condition and Pontryagin duality, $\text{Hom}(D,G)$ is LC for all LCA $G$.

Consider a discrete abelian group $G_0$ (I'll fix it later). For $F\subset D$ finite subset, define $U_F$ as the set of homomorphisms $D\to G_0$ vanishing on $F$. So the $U_F$ form a basis of clopen neighbourhoods of 0 in $\text{Hom}(D,G_0)$. So one of those, say $U_F$ is compact. If $E$ is the subgroup generated by $F$, this shows that $\text{Hom}(D/E,G_0)$ is compact. Now to specify, let us assume from the beginning we picked the group $G_0=H^{(\mathbf{N})}$, where $H=\mathbf{Q}/\mathbf{Z}$. It is a standard verification that if $D/E$ is nonzero, then $\text{Hom}(D/E,H)$ is not trivial, and an easy consequence is that $\text{Hom}(D/E,G_0)$ is not compact, a contradiction. This shows that $D/E=0$. So $D$ is finitely generated, and thus $T$ is a compact abelian Lie group, as required.

A careful look on the proof shows that otherwise, $\text{Hom}(G,T)$ is not locally compact for the specific choice of $G=\hat{\mathbf{Z}}^{\mathbf{N}}\times\mathbf{Z}^{(\mathbf{N})}$, where $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ and $\mathbf{Z}_p$ are the $p$-adics.

Edit: the answer for 2. is, as mentioned in another post, that only the circle satisfies the condition. (I initially thought it was trivially true for the 2-torus (or even the trivial group!), while it's trivially false.)

Here's a proof. Taking $G=\mathbf{R}/\mathbf{Z}$, the condition implies that $\text{Hom}(\mathbf{R}/\mathbf{Z},T)$ is nonzero; pick $f$; its kernel is finite and the quotient of $\mathbf{R}/\mathbf{Z}$ by the kernel of $f$ is also isomorphic to the circle, so the image is isomorphic to the circle. Now by Pontryagin duality, whenever $\mathbf{R}/\mathbf{Z}$ stands as a closed subgroup in a LCA group, it stands as a direct factor. So write $T=H\times \mathbf{R}/\mathbf{Z}$. Then for every $G$ we get $\text{Hom}(\text{Hom}(G,T),T)=G\times \text{Hom}(\hat{G},H)\times \widehat{\text{Hom}(G,H)}\times\text{Hom}(\text{Hom}(G,H),H)$. Since by assumption the identity of $G$ into the first factor is an isomorphism for every $G$, it follows that all other 3 factors are zero. Taking $G=\mathbf{Z}$, we deduce $\widehat{\text{Hom}(\mathbf{Z},H)}=0$, so $H=\text{Hom}(\mathbf{Z},H)=0$ by Pontryagin duality, hence $T=\mathbf{R}/\mathbf{Z}$.

any". If that example had already occurred to you, then you should have already said so in your question. – Todd Trimble♦ Nov 10 '12 at 14:52