(1) No. Example: $$ \begin{array}{ccc} & \alpha & \\ 1 & \rightleftarrows & 2 \\ & \beta & \end{array}$$ with all paths of length 3 = 0. This is a symmetric algebra wigh Loewy structure $$ \begin{array}{ccc} 1 & & 2\\ 2 & \oplus & 1 \\ 1 & & 2 \end{array}$$ and has infinite global dimension. Its Cartan matrix is $$ \left(\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right) $$ whose determinant is not 1, hence not inveritble in $\mathrm{Mat}_2(\mathbb{Z})$.
(2) In general, I don't know. But this is (partly) known for cellular algebras. (And I think BGG algebras, which always is of finite gloabl dimension, always have Cartan matrix with determinant 1, but I am not 100% sure, nor do I have any reference) See C.C.Xi's lecture notes. http://www.math.jussieu.fr/~keller/ictp2006/lecturenotes/xi.pdf The result is: the global dimension of a cellular algebra is finite if and only if the Cartan matrix has determinant 1.