Let $\tau^{*}$ the subspace euclidian topology over $[0,1]$
We define $\tau=\tau^{*} \cup${$\mathbb{R}$}.
(i) Are $\tau$ finer or coarser of the standard euclidian topology $\tau_e$?
(ii) Find the interior of $[0,+\infty)$, $[1/2,+\infty)$ and $\mathbb{Z}$
(iii)Find $\overline{[0,1]}$
(iv)Is $\mathbb{R},\tau$ an Hausdorff space?
(v) Is it compact? Connected?
(i) I think the two topologies are not comparable since $[0,1] \in \tau \setminus \tau_e$ and $(1,2) \in \tau_e \setminus \tau$.
(ii) I found $Int[0,+\infty)=[0,1]$.
$Int[1/2,+\infty)=(1/2,1]$
And $Int\mathbb{Z}=\emptyset$
(iii) $\overline{[0,1]}=\mathbb{R}$
(iv) This space is not $T_2$ because every $x \in [0,1]$ is not Hausdorff separable from the points $y \in (-\infty,0) \cup (1,+\infty)$. In fact every open neighborhood of $y$ is the whole $\mathbb{R}$, so there aren't disjoint neighborhoods for $x$ and $y$, and the space is not $T_2$.
(v) It's compact because for every open cover, {$\mathbb{R}$} is a finite open cover for $\mathbb{R}$
It's connected because there aren't open, non empty, sets such that they are a partition of $\mathbb{R}$
Handy and obvious fact (that you already mentioned as well):
This implies most of the following answers.
(i) is fine; (ii) are both correct, as no point outside $[0,1]$ be interior; (iii) is too, as all points outside $[0,1]$ are limit points of $[0,1]$; (iv) also implies that two points outside $[0,1]$ can never be separated.
As to (v): a little more argumentation would be nice:
$\mathbb{R}$ is compact, indeed, as for any open cover of $\mathbb{R}$, to cover $2$, e.g. must contain $\mathbb{R}$ as an element and then $\{\mathbb{R}\}$ is a very finite subcover.
connected: if $\mathbb{R} = U \cup V$ for open non-empty sets, then say $2 \in U$ and the fact implies that $U = \mathbb{R}$. So $U$ and $V$ are never disjoint. So $\mathbb{R}$ is connected in this topology.