The functions of each pair are inverse to each other. For each pair, check that both compositions give the identity function.
$F: \mathbb{R} \to \mathbb{R}$ and $F^{−1}:\mathbb{R} \to \mathbb{R}$ are defined by $F(x)=3x+2$ and $F^{−1}(y)=\dfrac{y−2}{3}$, for all $y \in \Bbb{R}.$
My attempt:
Inverse Function
For each particular but arbitrarily chosen $y \in \mathbb{R}$, according to the definition of $f^{-1}$, $f^{-1}(y) = \dfrac{y-2}{3}$ is a unique real number $x$ such that $f(x) = y$. \begin{align*} F(x) & = y\\ 3x + 2 & = y\\ x & = \frac{y-2}{3} \end{align*} Therefore, $f^{-1}(y) = \frac{y-2}{3}$.
Compositions of Functions.
The functions $g \circ f$ and $f \circ g$ are defined as follows: $$(g \circ f)(x) = g(f(x)) = g(3x + 2) = 3x + 2$$ for all $x \in \mathbb{Z}$.
If I understand well then you must check whether the functions $\mathbb R\to\mathbb R$ prescribed by $x\mapsto3x+2$ and $x\mapsto\frac13(x-2)$ are inverses of each other.
That is indeed the case since their compositions are prescribed by: $x\mapsto3x+2\mapsto\frac13((3x+2)-2)=x$ and: $x\mapsto\frac13(x-2)\mapsto3(\frac13(x-2))+2=x$ so both are identity functions.
This observation is enough to conclude that the functions are inverses of each other, and is what you are asked to do.