Roots of the Chebyshev polynomials of the second kind.

For your first case:

Since $U_n(x) =\frac{\sin((n+1)t)}{\sin(t)} $ where $x = \cos(t) $,

$\begin{array}\\ U_n(x)+U_{n-1}(x) &=\frac{\sin((n+1)t)}{\sin(t)}+\frac{\sin(nt)}{\sin(t)}\\ &=\frac{\sin((n+1)t)+\sin(nt)}{\sin(t)}\\ &=\frac{2\sin((n+1/2)t)\cos(t/2)}{\sin(t)}\\ \end{array} $

and this is zero when $t(n+1/2) =k\pi $ for some integer $k$, or $t =\frac{k\pi}{n+1/2} $ for $1 \le k \le n$.

This gives $n$ real roots, and that is all since $U_n(x)+U_{n-1}(x)$ is of degree $n$.