Answer: a) it is possible; b) it is impossible.

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 $2 \times 2$ corner squares:

All these squares have the same form and per one move (+1 - 1) they can be transformed into the squares with $1$ in all their cells. So we can obtain the following table:

Further, we transform the third row of the table:

In a similar way we can transform the third 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 $13$ white and $12$ 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 $\frac{1+25}{2} \cdot 13 = 13 \cdot 13$, and the sum $S_b$ of the numbers in the black cells is equal to $\frac{2+24}{2} \cdot 12 = 13 \cdot 12$. So, $S_w - S_b = 13$.
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 white and black cells. If all cells in the table are occupied with $2$, then this difference is equal to $2$. But $2 \not\equiv 13 \pmod{3}$, so we cannot obtain the table with $2$ in all its cells.