1) Does there exist $f(x)$ and $g(x)$ such that $f(g(x))=x$ but $f(x)$ and $g(x)$ are not inverse of each other.
2) Does $f(g(x)) = x$ automatically imply $g(f(x))=x$?
Note that an example can be fabricated as follows :
Let $g(x) = 1$ for $x = 0$ and it is only defined for $x=0$.
Let $f(x) = x-1$.
Clearly $f(g(x)) = x$ at $x = 0$ and $f(x)$ and $g(x)$ are not inverse of each other.
I am however looking for two continuous functions $f(x)$ and $g(x)$, not an isolated function as is $g(x)$ in the example above.
Given any surjective function $f:A \to B$, you can find such $g$ (if you assume the axiom of choice). The construction is as follows:
Consider for any fixed $b \in B$, define $K_b=\{a\in A: f(a)=b\}$. By surjectivity, this set is non-empty for all $b\in B$. Now, by the axiom of choice, we can choose an $a_b\in K_b$ for each $b\in B$. Then we can define $$g:B\to A, b\mapsto a_b$$ Note that for any $b\in B$, $f(g(b))=f(a_b)=b$.
Obviously, there are a lot of continuous surjective functions that are not injective.
Remark: In fact, you can also show that if $f:A\to B$ has a function $g:B\to A$ such that $f(g(x))=x$, then $f$ is surjective. (This part would not require the axiom of choice.)