We will count acute triangles $T_iT_jT_k$ with $1\le i<j<k\le n$ so that no two triangles are counted twice. Since $n$-gon is regular we can count the triangles with one fixed vertex, say $i=1$, and then multiply received number times $n$. Observe $$k\ge \left\lfloor \frac{n}{2}\right\rfloor+2\iff \angle T_iT_jT_k<90^\circ,$$$$ j\le \left\lfloor \frac{n+1}{2}\right\rfloor\iff\angle T_iT_kT_j<90^\circ,$$$$k-j\le \left\lfloor \frac{n-1}{2}\right\rfloor\iff \angle T_kT_iT_j<90^\circ$$Observe that all $$ k\in\left\lbrace \left\lfloor \frac{n}{2}\right\rfloor +2,\left\lfloor\frac{n}{2}\right\rfloor +3,...,n \right\rbrace,\ j\in\left\lbrace k-\left\lfloor \frac{n-1}{2}\right\rfloor ,k-\left\lfloor \frac{n-1}{2}\right\rfloor +1,...,\left\lfloor \frac{n+1}{2}\right\rfloor\right\rbrace$$work. Therefore the number of acute triangles with $i=1$ is equal to $$\sum_{k=\left\lfloor n/2\right\rfloor+2}^n\left(1+\left\lfloor \frac{n-1}{2}\right\rfloor+\left\lfloor \frac{n+1}{2}\right\rfloor-k\right)=\sum_{k=\left\lfloor n/2\right\rfloor+2}^n\left(2\left\lfloor \frac{n+1}{2}\right\rfloor-k\right)=\sum_{k=1}^{\left\lfloor (n-1)/2\right\rfloor} \left(2\left\lfloor \frac{n+1}{2}\right\rfloor-k-\left\lfloor \frac{n}{2}\right\rfloor-1\right)=$$$$=\frac{3n-2-5\left\lfloor \frac{n}{2}\right\rfloor}{2}\cdot\left\lfloor \frac{n-1}{2}\right\rfloor.$$What's the total number of acute triangles in regular $n$-gon? We must multiply the result by $n$ (since the vertice $1$ is freed now) and divide by $3$ because triangle has three vertices and therefore was counted three times. Finally the probability is $$P(A)=\frac{n\cdot\frac{3n-2-5\left\lfloor \frac{n}{2}\right\rfloor}{2}\cdot\left\lfloor \frac{n-1}{2}\right\rfloor}{3{n\choose3}}=\frac{3n-2-5\left\lfloor \frac{n}{2}\right\rfloor}{(n-2)(n-1)}\cdot\left\lfloor \frac{n-1}{2}\right\rfloor.\blacksquare$$

Homework for you: do the same for obtuse triangle.

In this solution, we will first fix a vertex, and name it $V_0$. We then name the vertices on one side as $A_1, \; A_2 \dots$ , and $B_1, \; B_2 \dots $ on the other side, until all vertices are named. Now, if a triangle containing $V_0$ is acute, then it must be of the form $A_iV_0B_j$, with some further conditions on $i$ and $j$ mentioned later. We now consider $2$ cases, where $n$ is odd and even. $\emph{Case 1}:$ When $n$ is odd. Say $n=2k+1$. In this case, the vertices are marked as $A_k, \; A_{k-1} \dots , \;A_1$, $V_0$, $B_1, \; B_2 \dots , \;B_k$. Now, if one of the chosen vertices is $A_i$, then there are $i$ possibilities for the other vertex. This is because if we reflect $A_i$ about the center of the circle, say at $C_i$, we want the number of points 'below' $C_i$, which is $1$ more than $C_{i-1}$, and for $C_1$ the number of points below it is $1$. Summing over all $i$, the total possiblities $= \sum\limits_{i=1}^k i= \frac{k(k+1)}{2}$. So, in this case, $P(acute)=\frac{\frac{k(k+1)}{2}}{{{2k}\choose{2}}}=\frac{k+1}{2(2k-1)}=\frac{n+1}{4(n-2)}$ $\emph{Case 2}:$ When $n$ is even. Say $n=2k$. In this case, the vertices are marked in such a way that $A_K$ is the same as $B_K$. We can just ignore this vertex, due to it's being opposite $V_0$, which would result in the formation of a right-angled triangle. Thus, we are only concerned with named $A_{k-1}, \; A_{k-2} \dots , \;A_1$, $V_0$, $B_1, \; B_2 \dots , \;B_{k-1}$. In this case, if one of the chosen vertices is $A_i$, then there are $i-1$ possibilities for the other vertex. The reasoning is similar as before, except that the $C_i$ in this scenario is $B_{k-i}$. Summing over all $i$, the total possiblities $= \sum\limits_{i=1}^{k-1} i-1= \frac{(k-2)(k-1)}{2}$. So, in this case, $P(acute)=\frac{ \frac{(k-2)(k-1)}{2}}{{{2k-1}\choose{2}}}=\frac{k-2}{2(2k-1)}=\frac{n-4}{4(n-1)}$