1) If the limit [tex]L[/tex] is
[tex]L = \displaystyle \lim_{\Delta x\to0} \frac{\sin\left(\frac\pi3 + \Delta x\right) - \sin\left(\frac\pi3\right)}{\Delta x}[/tex]
then using the hint as well as [tex]\sin\left(\frac\pi3\right)=\frac{\sqrt3}2[/tex] and [tex]\cos\left(\frac\pi3\right)=\frac12[/tex] we have
[tex]\displaystyle L = \lim_{\Delta x\to0} \frac{\sin\left(\frac\pi3\right)\cos(\Delta x) + \cos\left(\frac\pi3\right)\sin(\Delta x) - \sin\left(\frac\pi3\right)}{\Delta x}[/tex]
[tex]\displaystyle L = \sin\left(\frac\pi3\right) \lim_{\Delta x\to0} \frac{\cos(\Delta x) - 1}{\Delta x} + \cos\left(\frac\pi3\right) \lim_{\Delta x} \frac{\sin(\Delta x)}{\Delta x}[/tex]
[tex]L = \dfrac{\sqrt3}2 \times 0 + \dfrac12\times1 = \boxed{\dfrac12}[/tex]
which follows from the well-known limits,
[tex]\displaystyle \lim_{x\to0} \frac{1-\cos(x)}x = 0 \text{ and } \lim_{x\to0} \frac{\sin(x)}x=1[/tex]
Alternatively, if you already know about derivatives, we can identify the limit as the derivative of [tex]\sin(x)[/tex] at [tex]x=\frac\pi3[/tex], which is [tex]\cos\left(\frac\pi3\right)=\frac12[/tex].
2) It looks like you may be using double square brackets deliberately to denote the greatest integer or floor function which rounds the input down to the nearest integer. That is, [tex][\![x]\!][/tex] is the greatest integer that is less than or equal to [tex]x[/tex]. The existence of [tex]L[/tex] depends on the equality of the one-sided limits.
Suppose [tex]3\le x<4[/tex]. Then [tex]2[tex]\displaystyle \lim_{x\to4^-} [\![x - 1]\!] = 2[/tex]
Now suppose [tex]4\le x<5[/tex], so that [tex]3\le x-1 < 4 \implies [\![x-1]\!]=3[/tex] and
[tex]\displaystyle \lim_{x\to4^+} [\![x-1]\!] = 3[/tex]
The one-sided limits don't match so the limit doesn't exist.
