It's easy, believe me!

Firstly we factor the $lcm$

$x^3+4x^2+x-6=(x-1)(x+2)(x+3)$

So $x^2+a-2$ and $x^2+5x+b$ are each the product of $2$ of the $3$ factors of the $lcm$

In the first polynomial the products of the roots is $-2$ , so we must have

$x^2+ax-2=(x-1)(x+2)=x^2+x-2 \Longrightarrow \mathbin{\color{#D61F06}a=1}$

In the second one the sum of the roots is $-5$ , so we must have

$x^2+5x+b=(x+2)(x+3)=x^2+5x+6 \Longrightarrow \mathbin{\color{#3D99F6}b=6}$

$\Longrightarrow \mathbin{\color{#D61F06}a}+\mathbin{\color{#3D99F6}b}=\mathbin{\color{#D61F06}1}+\mathbin{\color{#3D99F6}6}=\boxed{7}$

factorize lcm by factor thoeorem (x-1)(x+2)(x+3) therefore 1 st is (x-1)(x+2)=x^2+x-2 2nd is (x+2)(x+3)=x^2+5x+6 a+b is 7