$$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align} &A = 3 - \frac {2}{A}\\ \implies &\frac {A^2-3A+2}{A}=0\\ \implies &A^2-3A+2=0\\ \implies &(A-1)\cdot(A-2)=0\\ \implies &A=1\;\text{ or }\; A=2 \end{align}$$
I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.
Let us define two series. The first is \begin{align} a_1 &= 3 \\ a_2 &= 3 - \frac{2}{3} \\ a_3 &= 3 - \frac{2}{3 - \frac{2}{3}} \\ a_4 &= 3- \frac{2}{3 - \frac{2}{3 - \frac{2}{3}}} \\ &\vdots \\ a_{n+1} &= 3 - \frac{2}{a_n} \quad (*) \end{align} and \begin{align} b_1 &= 3 - 2 \\ b_2 &= 3 - \frac{2}{3-2} \\ b_3 &= 3 - \frac{2}{3 - \frac{2}{3-2}} \\ b_4 &= 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{3-2}}} \\ &\vdots \\ b_{n+1} &= 3 - \frac{2}{b_n} \quad (**) \\ \end{align}
Note: This is the same recurrence relation $(*)$ or $(**)$ but with different start value $a_1 = 3$ and $b_1 = 1$.
For convergence we need $a_{n+1} - a_{n} \to 0$ or $a_n \to a$.
This way (in case of convergence) equations $(*)$ and $(**)$ have a limit $$ a = 3 - \frac{2}{a} \quad (\#) $$ which has indeed the solutions $a = 1$ and $a = 2$.
However that means we could also try $$ c_n = 3 - \frac{2}{c_{n+1}} $$ or $$ c_{n+1} = \frac{2}{3 - c_n} \quad (\#\#) $$ because it has the limit form $(\#)$.
Note that equation $(\#\#)$ is quite different from equation $(*)$ (see image below).
And indeed this recurrence relation $(\#\#)$ works too. Using $c_1 = 1$ will give $c_n \to 1$, Using $c_1 = 2$ will give $c_n \to 2$. Using $c_1 = 1000$ will give $c_n \to 1$.
So why is this? Still two solutions and the start value decides the limit.
Here is an image:
The green graph is related to $(*)$: $$ f(x) = 3-\frac{2}{x} $$ the blue graph is related to $(\#\#)$: $$ g(x) = \frac{2}{3-x} $$ and the red graph is the identity function: $$ \mbox{id}(x) = x $$
We see that both $f$ and $g$ hit the identity at $x=1$ and $x=2$. Those points are fixed points of $f$ and $g$: \begin{align} x^* &= f(x^*) \\ x^* &= g(x^*) \end{align} And one could now try to apply the theory of fixed points, esp. properties of fixed point iterations. \begin{align} x_{n+1} &= f(x_n) \quad (\$) \\ x_{n+1} &= g(x_n) \end{align}
The fixed point iteration of $f$ is like the the iteration of original continued fractions (compare $(\$)$ with $(*)$ or $(**)$).
The theory behind can now help with statements about convergence and the dependency of start values.
This continued fraction is the limit of the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.
Note that as we add more terms to the continued fraction, it oscillates between $1$ and slightly higher than $2$. $$ \begin{align} n&=1& 3&=3& 3-2&=1\\\\ n&=2& 3-\cfrac23&=\frac73& 3-\cfrac{2}{3-2}&=1\\\\ n&=3& 3-\cfrac{2}{3-\cfrac23}&=\frac{15}7& 3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\ n&=4& 3-\cfrac2{3-\cfrac2{3-\cfrac23}}&=\frac{31}{15}& 3-\cfrac2{3-\cfrac2{3-\cfrac2{3-2}}}&=1\\\\ && 3-\cfrac2{\cfrac{2^n-1}{2^{n-1}-1}}&=\frac{2^{n+1}-1}{2^n-1}& 3-\cfrac21&=1 \end{align} $$ Therefore, the limit of the continued fractions with $2n-1$ twos and threes is $$ \lim_{n\to\infty}\frac{2^{n+1}-1}{2^n-1}=2 $$ and the limit of the continued fractions with $2n$ twos and threes is $$ \lim_{n\to\infty}1=1 $$ Therefore, one value represents the limit of an odd number of twos and threes and the other value represents the limit of an even number of twos and threes.
Your two solutions are correct. You can verify them as follows.
$$\color{blue}{1} = 3 - 2=3-\frac{2}{\color{\red}{1}}\tag{1}$$ Now replace the red 1 with the blue 1, which equals to the right hand side in (1).
$$\color{blue}{2} = 3 - 1=3-\frac{\color{cyan}{2}}{\color{\red}{2}}\tag{2}$$ Now replace the red 2 with the blue 2, which equals to the right hand side in (2).
I heard from the internet that this is one of the ways Ramanujan created some of his equalities.
Now 2 questions for you, (1) what if you replace the cyan 2 with blue 2 and so on so forth? (2) what if you replace the cyan 2 with blue 2 and then replace the red 2 with blue 2 and so on so forth?
