How many L-values determine a modular form? Suppose $f$ and $g$ are two newforms of certain levels, weights etc. If we know that
L(f,n)=L(g,n) for all sufficiently large $n$, can we conclude that $f=g$? 
Same question when the forms have the same weight and $n$ runs over critical points.
 A: I think the answer to your first question is "yes."  Suppose $L(f,s) = \sum_{m} a(m)m^{-s}$ and $L(g,s) = \sum_{m} b(m) m^{-s}$, and that $L(f,n) = L(g,n)$ for $n \geq n_0$, with $n_0$ large enough that the sums converge absolutely.  Then pick an integer $M \geq n_0$ and weights $C_M(n)$ so that $\sum_{n \geq M} C_M(n) m^{-n}$ is $1$ if $m=M$, and $0$ otherwise.  One can surely come up with such weights without too much trouble.  Then $a(M) = \sum_{n \geq M} C_M(n) L(f,n) = \sum_{n \geq M} C_M(n) L(g,n) = b(M)$.  It's not too hard to see that if two modular forms eventually have the same Fourier coefficients, then they are the same.
edit: After some further thought, I'm having trouble justifying the existence of those weights.  I found a different solution that I'm posting as a separate answer.
A: The answer to the first question is "yes".  The standard proof of the uniqueness of a Dirichlet series expansion actually generalizes to show the following.
Theorem.  Suppose that $A(s) = \sum_n a_n n^{-s}$ and $B(s) = \sum_n b_n n^{-s}$ are Dirichlet series with coefficients $a_n, b_n$ bounded by a polynomial.  If there exists a sequence of complex numbers $s_k$ with real part approaching infinity such that $A(s_k) = B(s_k)$ for all $k$, then $a_n = b_n$ for all $n$.
Proof (sketch).  Proceed by induction.  For $k$ big we have $A(s_k) = a_1 + O(2^{-\sigma_k})$ where $\sigma_k$ is the real part of $s_k$.  Similarly, $B(s_k) = b_1 + O(2^{-\sigma_k})$.  Since $A(s_k) = B(s_k)$, we conclude that $a_1 = b_1$.  A similar argument shows $a_2 = b_2$, $a_3 = b_3$, etc.
A: I think the answer to your second question is "no". For example if $k=2$ and $f$ and $g$ correspond to elliptic curves over $Q$ with positive rank, then the only critical point is $s=1$ and (at least conjecturally, and in sufficiently many cases provably) both $L$-functions will vanish at this point.
