The correct answer is $\boxed{\text{(A) 5, 2}}$.
The output sequence $y[n]$ is given by the convolution of the input sequence $x[n]$ and the impulse response $h[n]$:
$$y[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]$$
In this case, we have:
$$y[0] = x[0] h[0] + x[2] h[-2] = 1 \cdot 1 + 1 \cdot 2 = 3$$
$$y[1] = x[0] h[1] + x[2] h[-1] = 1 \cdot -1 + 1 \cdot 0 = -1$$
$$y[2] = x[0] h[2] + x[2] h[0] = 1 \cdot 2 + 1 \cdot 1 = 3$$
$$y[n] = 0 \text{ for } n \neq 0, 1, 2$$
Therefore, the number of nonzero samples in the output sequence is 5, and the value of $y[2]$ is 3.
Option (B) is incorrect because the number of nonzero samples in the output sequence is 5, not 6.
Option (C) is incorrect because the value of $y[2]$ is 3, not 1.
Option (D) is incorrect because the number of nonzero samples in the output sequence is 5, not 3.