We denote the rows by $i = 1, \dots, n$ and the columns by $j = 1, \dots, n$ and let $a(i; j)$ be the number written in the square $(i; j)$.

a) No! Consider the table $A$ where the squares $(i; j)$ are mined if and only if $i \equiv j \equiv 1 \pmod 3$ and the table $B$ where the squares $(i; j)$ are mined if and only if $i \equiv j \equiv 2 \pmod 3$. Then all numbers written in the squares of $A$ and $B$ are equal to $1$ and therefore we can not determine which squares are mined.

b) Yes! We first determine the mined squares in the third row. It is easy to see that the number $b(j)$ of the mined squares amongst $(3; j-1)$, $(3; j)$, $(3; j+1)$ is equal to $a(2; j) - a(1; j)$. We now compare $b(1)$ and $b(2)$ and determine if the square $(3; 3)$ is mined. Next we compare $b(4)$ and $b(5)$ and decide if $(3; 6)$ is mined. The same argument shows which squares amongst $(3; 9)$, $(3; 12)$, ..., $(3; 2007)$ are mined.

We now compare $b(2007)$ and $b(2006)$ to decide if the square $(3; 2005)$ is mined, then compare $b(2004)$ and $b(2003)$ to see if $(3; 2002)$ is mined, etc. We finally know which squares amongst $(3; 1999)$, $(3; 1996)$, ..., $(3; 1)$ are mined. Now $b(1)$ shows if $(3; 2)$ is mined, then we compare $b(3)$ and $b(4)$ to determine if $(3; 5)$ is mined, etc., and find which squares amongst $(3; 8)$, $(3; 11)$, ..., $(3; 2006)$ are mined. Hence we know row $3$.

Analogously we can determine the mined squares in the rows $6$, $9$, $12$, $15$, ..., $2007$. Using similar argument we can determine the mined squares in the rows $2005$, $2002$, $1999$, ..., $4$, $1$ and the rows $2$, $5$, $8$, ..., $2006$.