An important point to note is to that in cylindrical coordinates, the unit vectors can be specified as:
$$\begin {align}
\mathbf{a_r}&=\cos\theta{\mathbf{i}} + \sin\theta{\mathbf{j}} \\
\mathbf{a_{\theta}}&=-\sin\theta{\mathbf{i}} + \cos\theta{\mathbf{j}} \\
\mathbf{a_z}&=\mathbf{k}
\end{align} \nonumber $$
Of course, you can refer to the literature on how to derive these unit vectors.
Now (and specific to your question), given that $\mathbf{A} = A_1\mathbf{a_r}+A_2\mathbf{a_{\theta}}$ then (with the component of $\mathbf{a_k}$ identically $0$), the Divergence of the vector $\mathbf{A}$ can be calculated as:
$$ \begin {align}
\nabla \cdot \mathbf{A} &= \left(\frac{\partial }{\partial{r}}\mathbf{a_r} + \frac{1}{r}\frac{\partial }{\partial{\theta}}\mathbf{a_{\theta}} \right) \cdot \left(A_1\mathbf{a_r}+A_2\mathbf{a_{\theta}} \right) \\
&= \mathbf{a_r}\cdot \frac{\partial }{\partial{r}} \left(A_1 \mathbf{a_r}\right) + \mathbf{a_r}\cdot \frac{\partial }{\partial{r}} \left(A_2 \mathbf{a_{\theta}}\right) \\
&= \frac{1}{r}\mathbf{a_{\theta}}\cdot \frac{\partial }{\partial{\theta}} \left(A_1 \mathbf{a_r}\right) + \frac{1}{r}\mathbf{a_{\theta}}\cdot \frac{\partial }{\partial{\theta}} \left(A_2 \mathbf{a_{\theta}}\right)
\end{align} \nonumber
$$
$$
\begin {align}
\Rightarrow \nabla \cdot \mathbf{A} &= \mathbf{a_r} \cdot \left(A_1 \frac{\partial }{\partial{r}}\mathbf{a_r} + \mathbf{a_r}\frac{\partial }{\partial{r}}A_1 \right) \\
&= \mathbf{a_r} \cdot \left(A_2 \frac{\partial }{\partial{r}}\mathbf{a_{\theta}} + \mathbf{a_{\theta}}\frac{\partial }{\partial{r}}A_2 \right) \\
&= \frac{1}{r}\mathbf{a_{\theta}} \cdot \left(A_1 \frac{\partial }{\partial{\theta}}\mathbf{a_r} + \mathbf{a_r}\frac{\partial }{\partial{\theta}}A_1 \right) \\
&= \frac{1}{r}\mathbf{a_{\theta}} \cdot \left(A_2 \frac{\partial }{\partial{\theta}}\mathbf{a_{\theta}} + \mathbf{a_{\theta}}\frac{\partial }{\partial{\theta}}A_2 \right)
\end{align} \nonumber
$$
Again, since $$\mathbf{a_r} =\cos\theta{\mathbf{i}} + \sin\theta{\mathbf{j}} $$ this implies that $\frac{\partial }{\partial{r}}\mathbf{a_r} = 0$ and $\frac{\partial }{\partial{\theta}}\mathbf{a_r} = -\sin\theta{\mathbf{i}} + \cos\theta{\mathbf{j}} = \mathbf{a_{\theta}}$.
Also, from
$$ \mathbf{a_{\theta}} =-\sin\theta{\mathbf{i}} + \cos\theta{\mathbf{j}} $$
this implies that
$\frac{\partial }{\partial{r}}\mathbf{a_{\theta}} = 0$ and $\frac{\partial }{\partial{\theta}}\mathbf{a_{\theta}} = -\cos\theta{\mathbf{i}} - \sin\theta{\mathbf{j}} = -\mathbf{a_r}$.
Substituting and simplifying gives:
$$
\begin {align}
\mathbf{a_r} \cdot \left(A_1 \frac{\partial }{\partial{r}}\mathbf{a_r} + \mathbf{a_r}\frac{\partial }{\partial{r}}A_1 \right) &= \frac{\partial }{\partial{r}}A_1 \\
\mathbf{a_r} \cdot \left(A_2 \frac{\partial }{\partial{r}}\mathbf{a_{\theta}} + \mathbf{a_{\theta}}\frac{\partial }{\partial{r}}A_2 \right) &= 0 \\
\frac{1}{r}\mathbf{a_{\theta}} \cdot \left(A_1 \frac{\partial }{\partial{\theta}}\mathbf{a_r} + \mathbf{a_r}\frac{\partial }{\partial{\theta}}A_1 \right) &= \frac{1}{r}A_1 \\
\frac{1}{r}\mathbf{a_{\theta}} \cdot \left(A_2 \frac{\partial }{\partial{\theta}}\mathbf{a_{\theta}} + \mathbf{a_{\theta}}\frac{\partial }{\partial{\theta}}A_2 \right) &= \frac{\partial }{\partial{\theta}}A_2
\end{align} \nonumber
$$
Therefore:
$$ \nabla \cdot \mathbf{A} = \frac{\partial }{\partial{r}}A_1 + \frac{1}{r}A_1 + \frac{\partial }{\partial{\theta}}A_2 $$
Now to put in a form that agrees with the given answer, notice that:
$$ \frac{1}{r} \frac{\partial }{\partial{r}}\left(rA_1\right) = \frac{1}{r} \left(r\frac{\partial}{\partial{r}}A_1 + A_1\frac{\partial}{\partial{r}}r \right) =\frac{\partial }{\partial{r}}A_1 + \frac{1}{r}A_1
$$
Summary:
$$
\begin {align}
\nabla \cdot \mathbf{A} &= \frac{\partial }{\partial{r}}A_1 + \frac{1}{r}A_1 + \frac{\partial }{\partial{\theta}}A_2 \\
&= \frac{1}{r} \frac{\partial }{\partial{r}}\left(rA_1\right) + \frac{\partial }{\partial{\theta}}A_2
\end{align} \nonumber
$$
Best Answer
Well, when you have $$ \frac{\partial}{\partial x_i}(\varepsilon_{ijk}u_jv_k), $$ you only wrote one term instead of two. Applying the product rule: $$ \varepsilon_{ijk}\frac{\partial u_j}{\partial x_i}v_k+\varepsilon_{ijk}u_j\frac{\partial v_k}{\partial x_i}=(\mathbf{\nabla}\times u)_kv_k-\varepsilon_{ikj}\frac{\partial v_k}{\partial x_i}u_j=(\mathbf{\nabla}\times u)_kv_k-(\mathbf{\nabla}\times v)_ju_j= $$ $$ =(\mathbf{\nabla}\times u)\cdot v-(\mathbf{\nabla}\times v)\cdot u. $$