1. We are given $f(x)=f^1(x)=\frac{3x+2}{2x+3}$ and $f^n(x)=f\circ f^{n-1}(x).$ We need to find $f^5(x)$ in the form $f^5(x)=\frac{ax+b}{bx+a},\quad \gcd(a,b)=1,$ and then compute $a+b$.

2. A linear fractional transformation $x\mapsto \frac{px+q}{rx+s}$ can be represented by the matrix $M=\begin{pmatrix}p&q\\ r&s\end{pmatrix}.$ Then composition corresponds to matrix multiplication.

Here, $f(x)=\frac{3x+2}{2x+3}$ corresponds to $M=\begin{pmatrix}3&2\\2&3\end{pmatrix}.$ Therefore, $f^n(x)\leftrightarrow M^n.$

3. Observe that $M=\begin{pmatrix}3&2\\2&3\end{pmatrix}$ is of the form $\begin{pmatrix}a&b\\b&a\end{pmatrix}.$ Such matrices remain of the same form under multiplication. So let $M^n=\begin{pmatrix}A_n&B_n\\B_n&A_n\end{pmatrix}.$ Then $f^n(x)=\frac{A_nx+B_n}{B_nx+A_n}.$

Thus, we only need $M^5$.

4. Diagonalize using the observation: For matrix $\begin{pmatrix}a&b\\b&a\end{pmatrix},$ the eigenvalues are $a+b$ and $a-b$.

For $M$: $3+2=5,\qquad 3-2=1.$ So eigenvalues are $5$ and $1$.

Hence for $M^5$, the corresponding eigenvalues are $5^5=3125,\qquad 1^5=1.$

If $M^5=\begin{pmatrix}A&B\\B&A\end{pmatrix},$ then its eigenvalues are $A+B\quad \text{and} \quad A-B.$ Therefore, $A+B=3125,\qquad A-B=1.$

Solving, $2A=3126 \implies A=1563,$ $2B=3124 \implies B=1562.$

So, $M^5=\begin{pmatrix}1563&1562\\1562&1563\end{pmatrix}.$ Thus, $f^5(x)=\frac{1563x+1562}{1562x+1563}.$

5. Therefore, $a=1563,\qquad b=1562.$ Since $\gcd(1563,1562)=1,$ the required sum is $a+b=1563+1562=3125.$

6. Comparison with stored correct answer: Stored answer = $3125$. Our derived answer = $3125$. They match.