Let $V$ be a vector space over $K$ and $v_1, ..., v_n \in V$. Consider those two statements:
(i) $\alpha_1v_1 + ... + \alpha_nv_n = 0$ only has the solution $\alpha_1 = ... = \alpha_n = 0$ (so the vectors $v_i$ are linearly independent)
(ii) There exist no coefficients $(\lambda_1, ..., \lambda_{n-1}) \neq (0,...,0)$, such that $v_n = \lambda_1v_1 + ... + \lambda_{n-1}v_{n-1}$
Show that (i) implies (ii), but (ii) does not imply (i).
(i) $\implies$ (ii)
Assume (i) holds, that is they are linearly independent. Then we can rewrite (ii) as $-\lambda_1v_1 - ... - \lambda_{n-1}v_{n-1} + v_n = 0$. If they are linearly independent, then we have $\lambda_1 = ... = \lambda_{n-1} = 0$
So we have $$0 + ... + 0 + v_n = 0$$
Is this correct until now ? What now ?
(ii) does not imply (i)
I suppose I must somehow find a counterexample, but I'm a bit lost about how.
Here is a bit of polishing on your proof, more in terms of style of writing.
Assume $v_n = \lambda_1v_1 + ... + \lambda_{n-1}v_{n-1}$. Then, you rewrite as $\lambda_1v_1 + ... + \lambda_{n-1}v_{n-1} + \lambda_n v_n =0$, where $\lambda_n= -1\neq0$. Then you have just found some $\alpha_1,...,\alpha_n$, not all zero, with $\alpha_1v_1 + ... + \alpha_nv_n = 0$ (namely $\alpha_i=\lambda_i$).
We have shown that the negation of (ii) implies the negation of (i). That is to say that (i) implies (ii).
For your second question, let's build a counterexample, and let's take $V=\mathbb{R}^2$ with $v_1=(1,0), v_2 = (2,0) $ and $v_3=(0,1)$. Clearly we have non independent vectors, and $v_3$ is not a linear combination of $v_1,v_2$. That is, (ii) holds, but not (i).