$(1)$
\[\begin{align}
F&=\rho Q\left(V-u\right)\\
&=\rho AV\left(V-u\right)
\end{align}\]
\[\begin{align}
L&=Fu\\
&=\rho AVu\left(V-u\right)
\end{align}\]
$(2)$
$\frac{dL}{du}=0$のとき$L$は最大となるので
\[\begin{align}
\frac{dL}{du}&=\frac{d\left\{\rho AVu\left(V-u\right)\right\}}{du}+\rho AV\left\{V-u+u\cdot\left(-1\right)\right\}\\
&=\rho AV\left(V-2u\right)
\end{align}\]
\[\underline{\therefore \rho AV\left(V-2u\right)=0}\]
$\rho AV\ne0$だから$V-2u=0 \ \Longrightarrow\ u=\frac{V}{2}$
\[
L_{max}=\rho AV\cdot\frac{V}{2}\left(V-\frac{V}{2}\right)=\frac{\rho AV^3}{4}\]
\[
\eta_{max}=\frac{最大動力値}{流体の持つ運動エネルギ}=\frac{\frac{\rho AV^{3}}{4}}{\frac{\rho AV^3}{2}}=\frac{1}{2}=0.5
\hspace{10mm}
\]
$(3)$
\[\begin{align}
F&=\rho Q\left(V-u-\left(-\left(V-u\right)\cos\beta\right)\right)\\
&= \rho Q\left(V-u\right)\left(1+\cos\beta\right)\\
&=\rho AV\left(V-u\right)\left(1+\cos\beta\right)
\end{align}\]
\[\begin{align}
L&=Fu\\
&=\rho AVu\left(V-u\right)\left(1+\cos\beta\right)
\end{align}\]