First we make some useful observations that will simplify the problem:

* The order in which the buttons are pressed is not relevant to the final state. We only care about the number of times each button was pressed.
* Since pressing the same button twice makes no changes, we can assume that each button was pressed 0 or 1 times.

a. The answer is no. We proceed by contradiction. Suppose that there is a way to turn every lamp on after pressing some buttons.
We denote by $a, b, c, d, e, f, g, h, i$ the number of times each button was pressed, as shown in the figure.
Since the upper left corner cell must be turned on, we know that $b+d$ must be odd. Also, since the lower right cell must be turned on, we have that $f+h$ must be odd. This means that $b+d+f+h$ must be even. On the other hand, since the central cell must be turned on, $b+d+f+h$ has to be odd, but this is a contradiction. Therefore, it is not possible to ensure that all lamps are turned on after pressing some buttons.

b. We start by observing that there are $2^9$ ways to press the buttons, since each of the nine buttons can be pressed 0 or 1 times. However, this does not mean that there are $2^9$ different board states that can be achieved, because there may be repetitions. So let's find out, for any given board state, how many times it appears among those $2^9$ results.
To do that, given a fixed board state, we need to count in how many ways we can press the buttons to return to the original board state at the end. Suppose that we press a sequence of buttons that returns to the original board state, where we name $a, b, c, d, e, f, g, h, i$ the number of times each button was pressed as we did in (a). For the corner cells to return to their original state, we need $b+d, d+h, h+f$, and $f+b$ to be even. This happens if and only if $b, d, h, f$ have the same parity, and since their possible values are 0 or 1, they must be equal. Note that this also implies that the central cell returns to its original state after pressing the buttons, since $b+d+h+f = 4b$ is even.
On the other hand, since the neighbors of the central cell must also return to their initial state, we need $a+c+e, c+e+i, g+e+i$ and $a+e+g$ to be even. Subtracting $(a+c+e)-(c+e+i) = a-i$ we get that $a$ has the same parity as $i$, so $a=i$. Similarly, we get $c=g$ and also that $e$ has the same parity as $a+c$.
We conclude that if we choose $a, b$ and $c$ arbitrarily, we can fix the other numbers to satisfy the conditions needed to return to the original board state after pressing the buttons, so every fixed board state appears $2^3$ times in the $2^9$ possible results after pressing buttons. Therefore, there are $\frac{2^9}{2^3} = 2^6$ different board states that can be achieved.