$$ k(x) =\begin{cases} \dfrac { ax+\sin2x }{ 3x } &\quad x>0 \\ a(x+1) &\quad x\le 0 \end{cases} $$
For $k$ is a continuous, I've found that a is equal to $1$ (correct?). Now, whether there are values of $a$ for which $k$ is a differentiable function. I am not quite sure that after differentiating $\frac{ax+\sin2x}{3x}$, then substitute zero to the formula, I got zero over zero. So, is there no value of $a$ for which $k$ is a differentiable function?
Thank you, god bless you.
On
Note that $$\frac{\sin 2x}{3x} = \frac{\sin 2x}{2x} \cdot \frac23 \xrightarrow[x\to0^+]{} 1\cdot\frac23=\frac23.$$
Therefore, $$k(x) = \frac{ax+\sin 2x}{3x} \xrightarrow[x\to0^+]{} \frac{a+2}3 \tag{*} \label1$$
If $k$ is continuous, $k(0) = a(0+1) = a= \lim_{x\to0^+}k(x)$, so $3a = a+2 \iff a=1$, confirming your claim that $a=1$. The given function becomes
$$ k(x) =\begin{cases} \dfrac13 + \dfrac{1}{3x}\sin2x &\quad x>0 \\ a(x+1) &\quad x\le 0. \end{cases} $$
To know the differentiability of $k$ at $x_0=0$, try the right-hand difference quotient $$\frac{k(x)-k(0)}{x-0} = \frac1x k(x) \underset{x\to0^+}{\sim} \frac1x$$ since \eqref{1} suggests that $k(x) \xrightarrow[x\to0^+]{} 1$, so $k(x)$ is bounded in a neighbourhood of $x_0=0$, but $\frac1x$ is unbounded there. Hence, we confirm that there's no $a \in \Bbb{R}$ such that $k$ is differentiable.
we have the $$\lim_{x\to 0}a(x+1)=a$$ and $$\lim_{x\to 0}\frac{ax+\sin(2x)}{3x}=\frac{2+a}{3}$$ so one can write $$\frac{ax+\sin(2x)}{3x}=\frac{a}{3}+\frac{\sin(2x)}{2x}\cdot \frac{2}{3}$$