I will solve the first case, and leave the second for the OP's self-study.
It is evident that $\theta >0$, and the problem is not trivial if $n>1$.
For the first case, the likelihood function is
$$L=\frac {1}{(4\theta)^n}\cdot I\{-\theta \leq x_{(1)} \leq x_{(n)} \leq 3\theta\}$$
We have two conditions here that must hold at the same time, for the likelihood to be non-zero. We can write
$$I\{-\theta \leq x_{(1)} \leq x_{(n)} \leq 3\theta\} = I\{-\theta \leq x_{(1)} \}\cdot I\{ x_{(n)} \leq 3\theta\}$$
Note that in principle the fact that the support extends to negative values does not guarantee that the available sample at hand will contain negative values also, or even that some of the values will be positive. So we are not guaranteed that the minimum order statistic will be negative, nor that the maximum order statistic will be positive.
So to cover all possible samples, the likelihood will be non-zero if $\theta \geq -x_{(1)}$ and $\theta \geq x_{(n)}/3 \implies \theta \geq \max \{-x_{(1)},x_{(n)}/3\}$.
At the same time, the likelihood under this condition is a decreasing function of $\theta$. So the best we can do is to select the minimum positive value for $\theta$ that at the same times ensures that the likelihood will be non-zero. So we obtain
$$\hat \theta_{MLE} = \max \{-x_{(1)},x_{(n)}/3\}$$
Then we have cases (I ignore for clarity the special cases of equality between the minimum and the maximum order statistic, which also includes the pathological case where both are equal to zero).
A) $x_{(1)} , x_{(n)} < 0$.
Here the only positive value and so the MLE is $\hat \theta = |x_{(1)}|$. The intuition is that if the maximum order statistic is negative, then the minimum order statistic will be closer to the unknown parameter.
B) $0< x_{(1)} , x_{(n)}$.
Here too we have only one positive candidate in the $\max$ condition, so $\hat \theta = x_{(n)}/3$
C) $ x_{(1)} < 0 < x_{(n)}$. Here both candidate values in the $\max$ condition will be positive so $\hat \theta = \max \{|x_{(1)}|,x_{(n)}/3\}$.
