$\text{Do enunciado temos que:}$ $\boxed{\lim\limits_{x \rightarrow 0}{\dfrac{\left( \sqrt[3]{ax+b}-2 \right)}{x}}=\dfrac{7}{12}}$ $\text{Aqui, utilizaremos uma propriedade de limites:}$ $\dfrac{\red {\lim\limits_{x \rightarrow 0}}{\left(\sqrt[3]{ax+b}-2\right)}}{\red {\lim\limits_{x \rightarrow 0}}{x}}=\dfrac{7}{12} \Rightarrow \red {\lim\limits_{x \rightarrow 0}}{\left(\sqrt[3]{ax+b}-2\right)}=\dfrac{7}{12}\cdot \red {\lim\limits_{x \rightarrow 0}}{x}$ $\text{Agora, analisamos a expressão para $x$ atigindo o limite:}$ $\left(\sqrt[3]{a\cdot \red {0}+b}-2\right)=\dfrac{7}{12}\cdot \red {0} \Rightarrow \sqrt[3]{b}-2=0 \boxed{\red{\therefore b=8}}$ $\text{Para encontrarmos o $\red{a}$ utilizaremos a regra de $\red{L'Hôpital}$, mas antes verifique:}$ $\dfrac{\left( \sqrt[3]{a\cdot \red{0}+8}-2 \right)}{\red{0}}=\red{\dfrac{0}{0}}$ $\lim\limits_{x \rightarrow 0}{\dfrac{\left( \sqrt[3]{ax+b}-2 \right)}{x}}=\lim\limits_{x \rightarrow 0}{\left( \dfrac{a}{3\cdot (ax+b)^{\frac{2}{3}}}\right) } \Rightarrow \dfrac{a}{3\cdot 8^{\frac{2}{3}}}=\dfrac{a}{3\cdot 4}=\dfrac{a}{12}=\dfrac{7}{12}$ $\boxed{\red{\therefore a=7}}$ $\text{Multiplica-se $\red{a}$ e $\red{b}$: $a\cdot b = 7\cdot 8= 56$}$