If $\lim _\limits{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda+\mu$ is equal to

Let $\quad f(x)= \begin{cases}(1+a x)^{1 / x} & , x<0 \\ 1+b, & x=0 \\ \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2}, & x>0\end{cases}$ be continuous at $x=0$. Then $e^a b c$ is equal to:

For $\alpha, \beta, \gamma \in \mathbf{R}$, if $\lim _\limits{x \rightarrow 0} \frac{x^2 \sin \alpha x+(\gamma-1) \mathrm{e}^{x^2}}{\sin 2 x-\beta x}=3$, then $\beta+\gamma-\alpha$ is equal to :