I think that below I manage to answer the first of your questions. I will be very grateful for verification of this argument. Unless it is correct, I will delete this answer.

I denote Lebesgue outer measure by $m^*$ and $\sigma$-algebra of Lebesgue measurable sets by $\mathcal{L}$. Let $\mathcal{B}(\mathbb{R})$ be the $\sigma$-algebra of Borel sets on $\mathbb{R}$ with respect to the usual topology. Suppose that $G$ is an additive subgroup of $\mathbb{R}$ such that $m^*(G)>0$.

**Lemma 1.**
*$G$ is dense in $\mathbb{R}$ with respect to the usual topology.*

**Proof.** In the proof of this lemma we consider $\mathbb{R}$ with the usual topology.

Suppose that for every $\epsilon >0$ there exists $g\in G$ such that $g\in (-\epsilon,\epsilon)\setminus \{0\}$. Then for every $x\in \mathbb{R}$ there exists $n\in \mathbb{Z}$ such that $|n\cdot g - x|<\epsilon$. Thus $G$ is dense in $\mathbb{R}$.

Now if $G$ is not dense in $\mathbb{R}$, then by what we said above, there must be $\epsilon>0$ such that $\{0\} = G\cap (-\epsilon, \epsilon)$. This means that $0$ is isolated in $G$. Since $G$ is a topological group, it is homogeneous (any two points have homeomorphic neighborhoods - just pick translation). We deduce that every point of $G$ is isolated and hence $G$ is a discrete subset of $\mathbb{R}$. A discrete subgroup of $\mathbb{R}$ is countable and hence $m^*(G)=0$. This is contradiction with $m^*(G)>0$.

**Lemma 2.**
*There exists a constant $c>0$ depending only on $G$ such that for every bounded interval $I\subseteq \mathbb{R}$ we have
$$m^*\left(G\cap I\right) = c\cdot m^*(I)$$*
**Proof.** Fix $a\in \mathbb{R}$ and consider a function $f_a:(a,+\infty)\rightarrow \mathbb{R}$ given by formula
$$f_a(x) = m^*\left((a,x]\cap G\right)$$
Next for any $h>0$ pick a function $D_{a,h}:(a,+\infty)\rightarrow \mathbb{R}$ given by formula
$$D_{a,h}(x)= f_a(x+h)-f_a(x) =m^*\left((a,x+h]\cap G\right) - m^*\left((a,x]\cap G\right) =$$
$$= m^*\left((x,x+h]\cap G\right)$$
The last equality follows from the Caratheodory's criterion. Since $G$ is an additive subgroup of $\mathbb{R}$ and Lebesgue outer measure is translation invariant, we derive that

$$D_{a,h}(x+g) = D_{a,h}(x)$$

for every $g\in G$ and $g>0$. By **Lemma 1** we know that $G_+ = \{g>0|g\in G\}$ is dense in $\mathbb{R}_+$. Now the fact that $D_{a,h}$ is continuous implies that

$$D_{a,h}(x+t) = D_{a,h}(x)$$

for every $t>0$. Thus $D_{a,h}$ is constant for every $h>0$ and this implies that monotone function $f_a$, which has derivative almost everywhere, has constant derivative everywhere. Say $f_a'(x) = c$ for every $x\in (a,+\infty)$. Next $c$ does not depend on $a<0$, because for $a_1,a_2$ functions $f_{a_1}$ and $f_{a_2}$ have the same slope on $\left(\max\{a_1,a_2\},+\infty\right)$. Moreover, $c>0$, since $m^*(G)>0$.

Now for given $a\in \mathbb{R}$ we have $f_a(a) = 0$ and $f_a'(x) = c$ for $x\in (a,+\infty)$. This implies that

$$m^*\left(G\cap (a,x]\right) = f_a(x) = c\cdot x - c\cdot a$$
for every $a\in \mathbb{R}$. If $I = (\xi_1,\xi_2]$ is an interval, then pick $a< \xi_1$ and then

$$m^*(G\cap I) = m^*(G\cap (\xi_1,\xi_2]) = m^*(G\cap (a,\xi_2]) - m^*(G\cap (a,\xi_1]) =$$
$$= f_a(\xi_2) - f_a(\xi_1) = \left(c\cdot \xi_2 - c\cdot a\right) - \left(c\cdot \xi_1 - c\cdot a\right) = c\cdot (\xi_2 - \xi_1) = c\cdot m^*(I) $$

Thus we proved that the statement of the lemma holds for intervals open from the left and closed from the right. Since every interval is up to (endpoints) a set of measure zero of the above form, we deduce that the result holds for all intervals.

**Lemma 3.**
*Let $E$ be a subset of $\mathbb{R}$ and let $c>0$. Suppose that $m^*(E\cap I) = c\cdot m^*(I)$ for every bounded interval $I\subseteq \mathbb{R}$. Then for every bounded set $A\in \mathcal{L}$ we have
$$m^*(E\cap A) = c\cdot m^*(A)$$*

**Sublemma.**
*Let $\{A_n\}_{n\in \mathbb{N}}$ be a family of pairwise disjoint members of $\mathcal{L}$ and let $E\subseteq \mathbb{R}$ be a bounded subset. Then
$$m^*\left(E\cap \bigcup_{n\in \mathbb{N}}A_n\right) = \sum_{n\in \mathbb{N}}m^*(E\cap A_n)$$*
**Proof of the sublemma.**
It is a consequence of the fact that $\mathcal{L}$ is constructed via Caratheodory's criterion that this result holds for finite family $\{A_n\}_{n=0}^N$ of pairwise disjoint sets in $\mathcal{L}$. We use this in the proof of the general case. We have
$$m^*(E) \leq m^*\left(E\cap \bigcup_{n\in \mathbb{N}}A_n\right) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right)\leq \sum_{n\in \mathbb{N}}m^*(E\cap A_n) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right) =$$
$$\leq \lim_{N\rightarrow +\infty}\left(\sum_{n=0}^Nm^*(E\cap A_n) + m^*\left(E\setminus \bigcup_{n=0}^NA_n\right)\right) =$$
$$ = \lim_{N\rightarrow +\infty}\left(m^*\left(E\cap \bigcup_{n=0}^NA_n\right) + m^*\left(E\setminus \bigcup_{n=0}^NA_n\right)\right)=\lim_{N\rightarrow +\infty}\mu^*(E) =\mu^*(E)$$
Thus in the inequality above, we have equality everywhere. In particular, we have
$$ m^*\left(E\cap \bigcup_{n\in \mathbb{N}}A_n\right) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right)= \sum_{n\in \mathbb{N}}m^*(E\cap A_n) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right)$$
The boundness of $E$ implies that we can cancel out $m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right) < +\infty$.

**Proof of the lemma 3.**
We prove that for every bounded interval $I$ and $A\in \mathcal{B}(\mathbb{R})$ such that $A\subseteq I$ we have

$$m^*(E\cap A) = c\cdot m^*(A)$$

Note that the family $\mathcal{F}$ of all such $A$ is a Dynkin system in the power set $\mathcal{P}(I)$. Indeed, if $A\in \mathcal{F}$ that is $m^*(E\cap A)= c\cdot m^*(A)$, then
$$c\cdot m^*(I) = m^*(E\cap I) = m^*(E\cap A) + m^*(E\cap (I\setminus A)) =$$
$$= c\cdot m^*(A) + m^*(E\cap (I\setminus A)) $$
Hence $m^*(E\cap (I\setminus A)) = c\cdot m^*(I\setminus A)$ and thus $I\setminus A\in \mathcal{F}$. Moreover, from **Sublemma** we derive that $\mathcal{F}$ is closed under countable unions of pairwise disjoint sets. By the assumption $\mathcal{F}$ contains all subintervals of $I$ and they form $\pi$-system. Now by Dynkin's $\pi\lambda$-theorem $\mathcal{F}$ contains all subsets in $\mathcal{B}(\mathbb{R})$ which are contained in $I$. Suppose that $A\in \mathcal{L}$ is contained in $I$, then $A = B\cup Z$, where $B\subseteq I$ and $B\in \mathcal{B}(\mathbb{R})$ and $m^*(Z)=0$. Hence

$$m^*(E\cap B) \leq m^*(E\cap A) = m^*\left((E\cap B)\cup (E\cap Z)\right) \leq m^*(E\cap B) + m^*(E\cap Z) = m^*(E\cap B) + 0 = m^*(E\cap B)$$

Thus $m^*(E\cap B) = m^*(E\cap A)$ and

$$m^*(E\cap A) = m^*(E\cap B) = c\cdot m^*(B) = c\cdot m(B) = c\cdot m(A) = c\cdot m^*(A)$$

This proves the lemma.

Now by **Lemma 2** and **Lemma 3**, we derive that there exists $c>0$ such that $m^*(G\cap A) = c\cdot m^*(A)$ for every bounded measurable subset $A$ of $\mathbb{R}$. Now pick any nonempty subset $A\in \mathcal{T}$. Intersect $A$ with some open interval $I$ such that $A\cap I\neq \emptyset$. Then $B = A\cap I\in \mathcal{T}$ is bounded. Thus
$$m^*(G\cap A) = m^*(G\cap B)= c\cdot m^*(B)>0$$
because $c>0$ and $m^*(B)>0$. Thus $G\cap A\neq \emptyset$. Therefore, $G$ is dense in $\mathcal{T}$.