Given the functions below, find $(g+h)(1)$.
\[
\begin{array}{l}
g(x)=x^{2}+4+2 x \\
h(x)=-3 x+2
\end{array}
\]
12
$-6$
6
8

Step 1 ：Given the functions \(g(x)=x^{2}+4+2x\) and \(h(x)=-3x+2\), we are asked to find the value of \((g+h)(1)\). This means we need to find the sum of the functions \(g(x)\) and \(h(x)\) at \(x=1\).

Step 2 ：First, we find the value of \(g(1)\) by substituting \(x=1\) into the function \(g(x)\). This gives us \(g(1)=1^{2}+4+2(1)=7\).

Step 3 ：Next, we find the value of \(h(1)\) by substituting \(x=1\) into the function \(h(x)\). This gives us \(h(1)=-3(1)+2=-1\).

Step 4 ：Finally, we add these two values together to get the final answer. This gives us \((g+h)(1)=g(1)+h(1)=7+(-1)=6\).

Step 5 ：So, the value of \((g+h)(1)\) is \(\boxed{6}\).