\(a,f\left(x\right):g\left(x\right)=\left(3x^4+9x^3+7x+2\right):\left(x+3\right)\\ =\left[3x^3\left(x+3\right)+7\left(x+3\right)-19\right]:\left(x+3\right)\\ =\left[\left(3x^3+7\right)\left(x+3\right)-19\right]:\left(x+3\right)\\ =3x^3+7.dư.19\)

\(c,\) Để \(k\left(x\right)⋮g\left(x\right)\Leftrightarrow-x^3-5x+2m=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow-\left(-3\right)^3-5\left(-3\right)+2m=0\\ \Leftrightarrow27+15+2m=0\\ \Leftrightarrow2m=-42\\ \Leftrightarrow m=-21\)

=>k^3+3k^2-k^2+9+6 chia hết cho k+3

=>\(k+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(k\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)

Thực hiện phép chia đa thức \(f\left(x\right)\)cho \(g\left(x\right)\)ta được: 

\(2x^3-3x^2+ax+b=\left(x^2-x+2\right)\left(2x-1\right)+\left(a-5\right)x+\left(b+2\right)\)

Để \(f\left(x\right)\)chia hết cho \(g\left(x\right)\)thì: 

\(\hept{\begin{cases}a-5=0\\b+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5\\b=-2\end{cases}}\).

Ta có: \(f\left(x\right)=x^4+mx^3+21x^2+x+n\)

\(=x^4-x^3-2x^2+\left(m+1\right)x^3-\left(m+1\right)x^2-\left(2m+2\right)x\) +(m+23)\(x^2\) -(m+23)x-(2m+46)+x(2m+2+m+23+1)+n+2m+46

\(=\left(x^2-x-2\right)\left(x^2+\left(m+1\right)\cdot x+m+23\right)\) +x(3m+26)+n+2m+46

Để f(x) chia hết cho g(x) thì 3m+26=0 và n+2m+46=0

=>3m=-26 và n=-2m-46

=>\(m=-\frac{26}{3};n=-2m-46=-2\cdot\frac{-26}{3}-46=\frac{52}{3}-46=\frac{-86}{3}\)