I think the answer to your first question is "yes." Suppose L(f,s)=summa(m)m−sL(f,s)=summa(m)m−s and L(g,s)=summb(m)m−sL(g,s)=summb(m)m−s, and that L(f,n)=L(g,n)L(f,n)=L(g,n) for ngeqn0ngeqn0, with n0n0 large enough that the sums converge absolutely. Then pick an integer Mgeqn0Mgeqn0 and weights CM(n)CM(n) so that sumngeqMCM(n)m−nsumngeqMCM(n)m−n is 11 if m=Mm=M, and 00 otherwise. One can surely come up with such weights without too much trouble. Then a(M)=sumngeqMCM(n)L(f,n)=sumngeqMCM(n)L(g,n)=b(M)a(M)=sumngeqMCM(n)L(f,n)=sumngeqMCM(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.
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