1. Let $f(x)=2\sqrt{3x+4}$. Find and simplify $f(-1),f(0),f(4),f(7)$, $f(-2),f(a),f(-1+h),f(c+h)$ if possible.

Step 1 ：Let's substitute the given values of x into the function $f(x)=2\sqrt{3x+4}$.

Step 2 ：For $f(-1)$, substitute $-1$ for $x$ in the function to get $f(-1)=2\sqrt{3(-1)+4}=2\sqrt{1}=2$.

Step 3 ：For $f(0)$, substitute $0$ for $x$ in the function to get $f(0)=2\sqrt{3(0)+4}=2\sqrt{4}=4$.

Step 4 ：For $f(4)$, substitute $4$ for $x$ in the function to get $f(4)=2\sqrt{3(4)+4}=2\sqrt{16}=8$.

Step 5 ：For $f(7)$, substitute $7$ for $x$ in the function to get $f(7)=2\sqrt{3(7)+4}=2\sqrt{25}=10$.

Step 6 ：For $f(-2)$, substitute $-2$ for $x$ in the function to get $f(-2)=2\sqrt{3(-2)+4}=2\sqrt{-2}=2\sqrt{2}i$.

Step 7 ：For $f(a)$, substitute $a$ for $x$ in the function to get $f(a)=2\sqrt{3a+4}$.

Step 8 ：For $f(-1+h)$, substitute $-1+h$ for $x$ in the function to get $f(-1+h)=2\sqrt{3(-1+h)+4}=2\sqrt{3h+1}$.

Step 9 ：For $f(c+h)$, substitute $c+h$ for $x$ in the function to get $f(c+h)=2\sqrt{3(c+h)+4}=2\sqrt{3c+3h+4}$.

Step 10 ：Final Answer: $f(-1)=\boxed{{\displaystyle 2}}$, $f(0)=\boxed{{\displaystyle 4}}$, $f(4)=\boxed{{\displaystyle 8}}$, $f(7)=\boxed{{\displaystyle 10}}$, $f(-2)=\boxed{{\displaystyle 2\sqrt{2}i}}$, $f(a)=\boxed{{\displaystyle 2\sqrt{3a+4}}}$, $f(-1+h)=\boxed{{\displaystyle 2\sqrt{3h+1}}}$, $f(c+h)=\boxed{{\displaystyle 2\sqrt{3c+3h+4}}}$