Let $g^{\prime}(1)=a$ and $g^{\prime \prime}(2)=b$

$\Rightarrow f(x)=x^{2}+a x+b$

Now, $f(1)=1+a+b ; f^{\prime}(x)=2 x+a ; f^{\prime \prime}(x)=2$

$g(x)=(1+a+b) x^{2}+x(2 x+a)+2$

$\Rightarrow g(x)=(a+b+3) x^{2}+a x+2$

$\Rightarrow g^{\prime}(x)=2 x(a+b+3)+a \Rightarrow g^{\prime}(1)=2(a+b+3)$$+a=a$

$\Rightarrow a+b+3=0$ .......(1)

$g^{\prime \prime}(x)=2(a+b+3)=b$

$\Rightarrow 2 a+b+6=0$ ........(2)

Solving (i) and (ii), we get

$a=-3$ and $b=0$

$f(x)=x^{2}-3 x$ and $g(x)=-3 x+2$

$f(4)=4$ and $g(4)=-12+2=-10$

$\Rightarrow f(4)-g(4)=16-2=14$