Why are V and I out of phase for a resistor in parallel with an inductor for the resistor?
\$V_{th}\$ is the voltage across \$R_L\$

No it isn't.
\$V_{TH}\$ is the open circuit voltage seen at the terminals of \$R_L\$ when \$R_L\$ is disconnected. It will be lower when \$R_L\$ is connected. It's called the Thevenin voltage of the source and it isn't the loaded value; it's the unloaded value.
In other words, the solver of the problem turned the 150 volt source, the 40 Ω resistor and the inductor and capacitor into a new source (\$V_{TH}\$) in series with \$Z_{TH}\$. This was done to the exclusion of \$R_L\$ because this is what you do when you transform a more complex source into a simpler source.
That new source becomes this: -



simulate this circuit &ndash; Schematic created using CircuitLab
And clearly, it has an open circuit voltage of \$V_{TH}\$ but will be loaded down when \$R_L\$ is connected.