Does Thevenin's theorem apply in power plants?
I know when we analyze a circuit's Thevenin equivalent, the best we
can ever get is 50% power transfer to the load.

That's a misunderstanding of the maximum power transfer theorem according to which the maximum power that can be delivered to the load is 50% of the maximum power available from the source.
But that doesn't mean that the best power ratio is 50%.
For a Thevenin voltage \$V_{th}\$, Thevenin resistance \$R_{th}\$ and load resistance \$R_L\$, the power delivered to the load is
$$P_L = \frac{V^2_{th}}{\left(R_{th} + R_L \right)}\frac{R_L}{\left(R_{th} + R_L \right)}$$
which is indeed maximum when \$R_L = R_{th}\$.
However, the power developed by the Thevenin source is
$$P_{th} =  \frac{V^2_{th}}{\left(R_{th} + R_L \right)}$$
and so the fraction of the source power delivered to the load is thus
$$\frac{P_L}{P_{th}} = \frac{R_L}{\left(R_{th} + R_L \right)}$$
So, for \$R_L &gt;&gt; R_{th}\$, almost all of the power supplied by the source is delivered to the load (though this power is much less than the maximum power available from the source).
I don't have the expertise to address the reasons for 62 units of energy lost in the power plant so this answer is just to address the proper interpretation of the maximum power transfer theorem.