Let $ABC$ be a right triangle where $\measuredangle A = 90^\circ$ and $M\in (AB)$ such that $\frac{AM}{MB}=3\sqrt{3}-4$. It is known that the symmetric point of $M$with respect to the line $GI$ lies on $AC$. Find the measure of $\measuredangle B$.
1. Given that $\triangle ABC$ is a right triangle with $\angle A = 90^\circ$, and $M \in AB$ such that $\frac{AM}{MB} = 3\sqrt{3} - 4$.
2. We need to find the measure of $\angle B$ given that the symmetric point of $M$ with respect to the line $GI$ lies on $AC$.

First, let's express $\frac{AM}{AB}$:
\[
\frac{AM}{AB} = \frac{AM}{AM + MB} = \frac{3\sqrt{3} - 4}{3\sqrt{3} - 4 + 1} = \frac{3\sqrt{3} - 4}{3\sqrt{3} - 3} = \frac{3(\sqrt{3} - 1) - 1}{3(\sqrt{3} - 1)} = \frac{3\sqrt{3} - 3 - 1}{3\sqrt{3} - 3} = \frac{3\sqrt{3} - 4}{3\sqrt{3} - 3}
\]

Simplify the fraction:
\[
\frac{AM}{AB} = \frac{5 - \sqrt{3}}{6}
\]

Let $K$ be the midpoint of $GI$ and $N \in AC$ be the reflection of $M$ in $K$. Since $\triangle AMN$ is a right triangle, the midpoint $K$ of its hypotenuse $MN$ is its circumcenter, so $AK = MK$.

Let $P$ be the midpoint of $AM$. Then:
\[
AP = \frac{AM}{2}
\]

Given that $r$ is the inradius of the right $\triangle ABC$, we have:
\[
r = \frac{bc}{a + b + c}
\]

Since $AP = \frac{AM}{2}$, we can write:
\[
AP = \frac{\frac{c}{3} + r}{2}
\]

We know:
\[
\frac{AM}{c} = \frac{1}{3} + \frac{b}{a + b + c} = \frac{5 - \sqrt{3}}{6}
\]

Thus:
\[
\frac{b}{a + b + c} = \frac{3 - \sqrt{3}}{6}
\]

Solving for $x$ in the equation:
\[
x + 1 + \sqrt{x^2 - 1} = \frac{a + b + c}{b} = 3 + \sqrt{3}
\]

where $x = \frac{a}{b} = \frac{1}{\sin B}$, we get:
\[
x = 2
\]

Thus:
\[
\sin B = \frac{1}{2}
\]

Therefore:
\[
B = 30^\circ
\]

The final answer is $\boxed{30^\circ}$