Answer
$L(x)=1+x$ and $Q(x)=1+x+\dfrac{x^2}{2!}$
Work Step by Step
Differentiate the given function $f(x)$ as follows:
$f'(x)=-\cos x e^{\sin x} ; \\f''(x)=-(\sin x) e^{\sin x}+(\cos x)^2 e^{\sin x} $
Now, $ f(0)=1; \\ f'(0)=1;\\ f''(0)=1$
$L(x)=f(0)+xf'(0)=1+x; \\ Q(x) =f(0) x +xf'(0) +\dfrac{x^2 f''(0)}{2!}=1+x+\dfrac{x^2}{2!}$
Our result is: $L(x)=1+x \\ Q(x)=1+x+\dfrac{x^2}{2!}$