No.

$\varprojlim_{n\in\mathbb{N}}\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}_p$, the $p$-adic integers.

Take $I=\mathbb{Q}_p$, the $p$-adic rationals.

There is a nonzero map $\mathbb{Z}_p\to\mathbb{Q}_p$, the inclusion, but $\operatorname{Hom}(\mathbb{Z}/p^n\mathbb{Z},\mathbb{Q}_p)=0$ for every $n\in\mathbb{N}$, since $\mathbb{Q}_p$ is torsion-free.

No.

$\varprojlim_{n\in\mathbb{N}}\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}_p$, the $p$-adic integers.

Take $I=\mathbb{Q}_p$, the $p$-adic rationals.

There is a nonzero map $\mathbb{Z}_p\to\mathbb{Q}_p$, the inclusion, but $\operatorname{Hom}(\mathbb{Z}/p^n\mathbb{Z},\mathbb{Q}_p)=0$ for every $n\in\mathbb{N}$, since $\mathbb{Q}_p$ is torsion-free.

Or alternatively, take $G_n=\mathbb{Z}$, with $G_{n+1}\to G_n$ multiplication by two. Then $\varprojlim G_n=0$ but $\varinjlim\operatorname{Hom}(G_n,\mathbb{Q})\cong\mathbb{Q}$.