a) We show that using the allowed moves we can decrease any number in the table by $3$ so that all other numbers in the table keep their values. We will not consider the whole table but only the number $x$ which will be decreased by $3$ and three more numbers $a$, $b$, and $c$ which occupy (together with $x$) the cells of some $2 \times 2$ square).

So we can consecutively decrease all numbers in the table so that to obtain their residues modulo $3$, i.e. to obtain the table:

Consider the $6 \times 6$ down-left corner square. It consists of the same four $3 \times 3$ squares:

It is easy to see that after two moves ($-1+1+1$) each of these $3 \times 3$ squares is transformed into the square with $1$'s in all its cells. So we obtain the table:

<table>
<tr>
<td>1</td>
<td>2</td>
<td>0</td>
<td>1</td>
<td>2</td>
<td>0</td>
<td>1</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>2</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>2</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
</tr>
</table>

Further, we transform the first row of the table:

$$\text{1 2 0 1 2 0 1 1 1 1 0 2 0 1 1 1 1 1 1 1 1}$$

In a similar way we can transform the last column of the table. As the result we obtain the table with $1$ in all its cells.

b) Consider the chess coloring of the table. For the definiteness we suppose that the corner cells are white. So, there are $25$ white and $24$ black cells in the table. All white cells are occupied with odd numbers, and all black cells are occupied with even numbers. Therefore, the sum $S_w$ of the numbers in the white cells is equal to $\frac{1+49}{2} \cdot 25 = 25 \cdot 25$, and the sum $S_b$ of the numbers in the black cells is equal to $\frac{2+48}{2} \cdot 24 = 25 \cdot 24$. So, $S_w - S_b = 25$.

It is easy to see that any allowed move does not change the residue modulo $3$ of the difference between the sums of the numbers in the white and black cells. If all cells in the table are occupied with the number $2013$, then this difference is equal to $2013$. But $2013 \ne 25 \pmod{3}$, so we cannot obtain the table with $2013$ in all its cells.