Proving decidability
For $L_1$, consider a computation $T(w)$ of a Turing machine $T$ that never moves its head to the left with input $w$. Let $\Gamma$ be the tape alphabet (including the blank symbol) and $Q$ be the set of states of $T$.
Notice that, at any given step during the computation, the future behavior of $T(w)$ is completely determined by:

The current state.
The tape symbol in the current cell.
The contents of the (infinitely many) tape cells on the right of the head.

There are $|Q|$ choices for 1, $|\Gamma|$ choices for $2$, and $\max\{|w|,1\}$ choices for 3 (since the contents of the tape to the right of the head are always a suffix of $w$ distinct from $w$, followed by infinitely many blank symbols).
To decide $L_1$ it suffices to simulate $T(w)$ for $\eta = |Q| \cdot |\Gamma| \cdot \max\{|w|,1\} + 1$ steps. If $T(w)$ moves its head to the left during the simulation, then reject. Otherwise accept. To see that accepting is the correct choice notice that if $T(w)$ does not terminate within $\eta$ steps, then a state must be repeated at least twice, implying that $T(w)$ is in some loop. This loop can only involve states that appeared at least once during the first $\eta$ steps, showing that $T$'s head will never move to the left.
For $L_2$ see this question.