For part b, I think YES.
can take $4$ values:
- $|f(x)-g(y)|$ and
For the first $2$ cases, $L1$ and $L2$ are the required Lipschitz constants, where
$|f(x)-f(y)| \le L1 |x-y|$
$|g(x)-g(y)|\le L2 |x-y|$
Now, for the $3$rd case, wlog, assume that $x < y$.
Then we must have $f(x)\ge g(x)$, but $f(y) \le g(y)$, so by the intermediate value theorem for the continuous function $f-g$, we get a point t such that $x < t < y$, satisfying: $f(t)=g(t)$.
$$|f(x)-g(y)| \le |f(x)-f(t)|+|f(t)-g(t)|+|g(t)-g(y)|\le
> L1(t-x)+L2(y-t) \le (\max(L1,L2))*(y-x)$$
The $4$th case is similarly dealt with and $\max(L1,L2)$ is your required Lipschitz constant for part b. Part a is all right, $L1*L2$ being the Lipschitz constant. Please correct me if I'm mistaken anywhere.