1) $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{uniform})$ is continuous.
Let $\langle x_n\rangle$ be a sequence in $\mathbb{R}$ and let $x\in \mathbb{R}$, and suppose $x_n\to x$ in $\mathbb{R}$ with the usual topology. It suffices to show that $h(x_n)\to h(x)$ in $(\mathbb{R}^\omega,\mbox{uniform})$. Since $h(x_n)$ and $h(x)$ are real valued functions, $h(x_n)$ converges to $h(x)$ uniformly iff $\sup_{k\in\omega}|h(x_n)(k)-h(x)(k)|\to 0$ in $\mathbb{R}$.
$\sup_{k\in\omega}|h(x_n)(k)-h(x)(k)|=\sup_{k\in\omega}|x_n/k-x/k|=|x_n-x|\cdot\sup_{k\in\omega}1/k=|x_n-x|\to 0$ since $x_n\to x$ in $\mathbb{R}$ by assumption.
2) $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{box})$ isn't continuous.
The box topology part is a bit more tedious. Let $x_n$ be a sequence in $\mathbb{R}$ which converges to $x$ in $\mathbb{R}$ with the usual topology. Then $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{box})$ is continuous iff
$\forall\epsilon_1,\epsilon_2,\dots>0,\exists N,\forall n\geq N,\forall k\in\omega,|h(x_n)(k)-h(x)(k)|<\epsilon_k$.
Observe $|h(x_n)(k)-h(x)(k)|=|x_n/k-x/k|=|x_n-x|/k<\epsilon_k$; taking $\epsilon_k=1/k^2$ our expression becomes $\exists N,\forall n\geq N,\forall k\in\omega,|x_n-x|<1/k$, or equivalently $\exists N,\forall n\geq N,|x_n-x|\leq\inf_{k\in\omega}1/k=0$. Choosing a sequence $\langle x_n\rangle$ in $\mathbb{R}$ which isn't eventually constant shows that $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{box})$ isn't continuous.
3) $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{product})$ is continuous. [I'm posting this one up since there have been some ugly proofs.]
Again let $x_n\to x$ in $\mathbb{R}$, then it suffices to show that $h(x_n)\to h(x)$ in $(\mathbb{R}^\omega,\mbox{product})$, which holds, iff $h(x_n)(k)\to h(x)(k)$ in $\mathbb{R}$ for all $k\in\omega$. Finally $h(x_n)(k)=x_n/k\to x/k=h(x)(k)$, and we are done.
We can use sequences since $\mathbb{R}$ is first countable.