Note that these questions are the same as asking whether non-zero $a_i$, $b_i$ can satisfy $\sum_{k=1}^\infty a_kx^{b_k}=0$ for all $x$ (or for x>1). This assumes the series converge absolutely, so that we can combine them. If we're not assuming absolute convergence, then can prove absolute convergence assuming that the $b_i$ are positive (this can be relaxed a bit, but not done away with).
Assuming the $b_i$ are strictly decreasing, we can answer it as follows. Assume the series converges absolutely for $x\ge A$ (that such an $A> 1$ exists will be shown at the end). Without loss of generality, assume $a_1\ne 0$. For $x\ge A$, we have $0=a_1 x^{b_1}+\sum_{n=2}^\infty a_nx^{b_n}$ So we solve for $a_1$ and take absolute values $|a_1|\le x^{-b_1}\sum_{n=2}^\infty |a_n|x^{b_n}=x^{-b_1+b_2}\sum_{n=2}^\infty |a_n|x^{b_n-b_2}$ Since the $b_i$ are strictly decreasing, both $-b_1+b_2$ and $b_n-b_2$ are negative (or zero, for $n=2$), thus for $x\ge A$, $x^{b_n-b_2}\le A^{b_n-b_2}$. Therefore, for $x\ge A$, $|a_1|\le x^{-b_1+b_2}\sum_{n=2}^\infty |a_n|A^{b_n-b_2}$ Since $x$ can be arbitrarily large, this implies that $a_1=0$. Contradiction.
We now prove the existence of such an $A$. Assume the series converges at $x_0$. Fix $\epsilon>0$ and set $x_n=x_0e^{\log(n^{-(1+\epsilon)})/b_n}$ Then $|a_n|x_n^{b_n}=|a_n|x_0^{b_n}(x_n/x_0)^{b_n}\le (|a_n|x_0^{b_n})n^{-(1+\epsilon)}$ Set $A=\lim\sup_n x_n$. Since the series converges at $x_0$, the terms $|a_n|x_0^{b_n}$ are bounded, hence the series will converge absolutely for $x\ge A$. A priori, $A$ could be infinite. But this won't happen as the $b_n$ are decreasing and positive.