field theory - If $L_1/K$ and $L_2/K$ are not Galois (solvable), then $L_1L_2/K$ is not Galois (solvable)

This is part of an exam preparation:

Prove/contradict:

1. If $L_1/K$ and $L_2/K$ are not Galois, then $L_1L_2/K$ is not Galois.


2. If $L_1/K$ and $L_2/K$ are not solvable Galois extensions, then $L_1L_2/K$ is not a solvable Galois extension.

For 1, I think that the answer is false. I thought of "splitting" the splitting field of a polynomial. For example, the splitting field of $X^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}(2^{1/3},\omega)$ and then taking $\mathbb{Q}(2^{1/3})$ and $\mathbb{Q}(\omega)$. The problem is that $\mathbb{Q}(w)/\mathbb{Q}$ is Galois and I can't think of a different kinds of example.

For 2, I think that the answer is true. We have

$$G_1 = Gal(L_1L_2/K) \leq Gal(L_1/K) \times Gal(L_2/K) = G_2$$

Then I thought of making some claim about a subgup of a solvable group but Im not really sure how to proceed.

Thanks.