Let $$f(x)=\int_0^x\left(t+\sin \left(1-e^t\right)\right) d t, x \in \mathbb{R}$$. Then, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}$$ is equal to

If the value of the integral $$\int_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$$ is $$\frac{2}{\pi}$$.Then, a value of $$\alpha$$ is

$$\text { Let } f(x)=\left\{\begin{array}{lr} -2, & -2 \leq x \leq 0 \\ x-2, & 0< x \leq 2 \end{array} \text { and } \mathrm{h}(x)=f(|x|)+|f(x)| \text {. Then } \int_\limits{-2}^2 \mathrm{~h}(x) \mathrm{d} x\right. \text { is equal to: }$$