$1.\left\{q\in \mathbb Q:1\le q^2\lt 3\right\}$
$1.\left\{q\in \mathbb Q:1\le q^2\le 3\right\}$
If the sets were simply $$1.\{r\in \mathbb R:1\le r^2\lt3 \}$$and$$2.\{r\in \mathbb R:1\le r^2\le3\}$$ we could say the first one is semi closed and the second one is compact. But the sets we happen to have here are $$1.\{r\in \mathbb R:1\le r^2\lt3 \}\cap\mathbb Q\\\text{and} \\ 2.\{r\in \mathbb R:1\le r^2\le3\}\cap \mathbb Q$$
So I think we need to consider two different cases::
Case $1.$
When our entire space is $\mathbb Q$ with subspace topology induced from $\mathbb R.$
In this case, both of the sets are closed and compact. Replacing $\lt$ by the $\le$ does not change anything in the given context.
We can rewrite the sets as follows-
$1.\mathcal A=\left\{q\in \mathbb Q:1\le q\lt \sqrt3\right\}$
$1.\mathcal B=\left\{q\in \mathbb Q:1\le q\le \sqrt3\right\}$
Now,$\sqrt3\notin\mathbb Q$
So,$\mathcal A^c=\left((-\infty,1)\bigcup (\sqrt3,\infty)\right)\cap \mathbb Q=\mathcal B^c$ which is open in the subspace topology that $\mathbb Q$ inherites from $\mathbb R$.And hence $\mathcal A$ and $\mathcal B$ are closed and bounded as well $\implies$ compact.
Case $2$:
When our entire space is $\mathbb R$ and we consider the sets $\mathcal A=\left\{q\in \mathbb Q:1\le q\lt \sqrt3\right\}$ and $\mathcal B=\left\{q\in \mathbb Q:1\le q\le \sqrt3\right\}$ above are considered as merely subsets of $\mathbb R$ to be judged then both sets are bounded but neither is even closed because neither contains the set of their limit points,i.e. the sets $\mathcal A \cap \mathbb Q^c$ for $\mathcal A$ and $\mathcal B\cap\mathbb Q^c$ of $\mathcal B.$
So,in both the cases, the signs $\le$ and $\lt$ do not make any difference in the results.
But when the case is not specified, I'm simply to assume that our space is $\mathbb R$ and the Case $2$ holds. Thoughts please. Thank You.