Let $b, g, r$ denote, respectively, the numbers on the blue, green, and red cards. Then we want to find the number of solutions of the equation
$$b + g + r = 16 \tag{1}$$
subject to the restrictions $3 \leq b \leq 9$, $3 \leq g \leq 7$, and $4 \leq r \leq 8$.
We can convert this to the equivalent problem in the nonnegative integers. Let $b' = b - 3$, $g' = g - 3$, and $r' = r - 4$. Then $b'$, $g'$, and $r'$ are nonnegative integers satisfying $b' \leq 6$, $g' \leq 4$, $r' \leq 4$. Substituting $b' + 3$ for $b$, $g' + 3$ for $g$, and $r' + 4$ for $r$ in equation 1 yields
\begin{align*}
b' + 3 + g' + 3 + r' + 4 & = 16\\
b' + g' + r' & = 6 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of $3 - 1 = 2$ addition signs in a row of six ones. For instance,
$$1 1 1 + 1 1 + 1$$
corresponds to the solution $b' = 3, g' = 2, r' = 1$ of equation 2 and $b = 6, g = 5, r = 5$ of equation 1, while
$$+ 1 1 + 1 1 1 1$$
corresponds to the solution $b' = 0, g' = 2, r' = 4$ of equation 2 and $b = 3, g = 5, r = 8$ of equation 1. The number of solutions of equation 2 in the nonnegative integers is the number of ways we can insert $3 - 1 = 2$ addition signs in a row of $6$ ones, which is
$$\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}$$
since we must choose which $2$ of the $8$ positions required for six ones and two addition signs will be filled with addition signs.
However, these solutions include those that violate the restrictions $g' \leq 4$ or $r' \leq 4$. Notice that both restrictions cannot be violated simultaneously since $2 \cdot 5 > 6$.
There are two ways to select the variable which exceeds $4$. Suppose it is $g'$. Then $g'' = g' - 5$ is a nonnegative integer. Substituting $g'' + 5$ for $g'$ in equation 2 yields
\begin{align*}
b' + g'' + 5 + r' & = 6\\
b' + g'' + r' & = 1 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{1 + 3 - 1}{3 - 1} = \binom{3}{2}$$
solutions. Hence, there are
$$\binom{2}{1}\binom{3}{2}$$
solutions of equation 2 which violate the restriction $g' \leq 4$ or $r' \leq 4$.
Therefore, the number of admissible solutions of equation 2 is
$$\binom{8}{2} - \binom{2}{1}\binom{3}{2} = 22$$
as you found.