Let $\alpha ,\beta$ are the roots of the equation $x^2+ax+b=0$ and $\gamma ,\delta$ are the roots of $x^2-ax+b-2=0.$ If $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }+\frac{1}{\delta }=\frac{5}{12}$ and $\alpha \beta \gamma \delta =24,$ then:

$\begin{array}{l}{x}^2+ax+b=0\begin{array}{c}\alpha \\ \beta \end{array}Also{x}^2-ax+b-2=0\begin{array}{c}\gamma \\ \delta \end{array}\\ b{x}^2+ax+1=0\begin{array}{c}1/\alpha \\ 1/\beta \end{array}(b-2)x^2-ax+1=0\begin{array}{c}1/\gamma \\ 1/\delta \end{array}\\ \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }+\frac{1}{\delta }=-\frac{a}{b}+\frac{a}{b-2}=\frac{5}{12}\\ \alpha \beta \gamma \delta =b(b-2)=24\\ \Rightarrow (b-6)(b+4)=0\\ \Rightarrow b=6,-4\\ Also,-a(\frac{1}{b}-\frac{1}{b-2})=\frac{5}{12}\\ -a\left(\frac{-2}{24}\right)=\frac{5}{12}\Rightarrow a=5\end{array}$