A1  ^?? Reduce the system of two differential equations for two amplitudes $a$ and $b$ to a single equation for amplitude $a$.

If we take the derivative of the first equation, then $a'‘ - i \kappa b’ = 0$, and we can substitute the second equation into the first to obtain
\[a''-\kappa^2 a = 0\]

A2  ^?? What boundary conditions are there for $a(0)$, $a(L)$, $a'(0)$ и $a'(L)$?

Remark: The existence of boundary conditions isn't necessary for all given values.

Boundary conditions are $a_0 = a(0)$ and $b(L)=0$. From the second we have $a'(L)=0$.

A3  ^?? Obtain the value of the reflection coefficient $R$.

The solutions to the equation for $a(x)$ are:
\[a(x) = Ae^{\kappa x} + B e^{-\kappa x}.\]The boundary conditions constrain the constants as follow: $A+B=a_0$ and $Ae^{\kappa L}-Be^{-\kappa L}=0$. Then, we have
\[ A = \frac{a_0}{1+e^{2\kappa L}}, \quad B = \frac{a_0}{1 + e^{-2\kappa L}}\]With $b(0) = -ia'(0)/\kappa$ we have
\[ b(0)=i(B-A)=ia_0 \left( \frac{1}{1+e^{-2\kappa L}} - \frac{1}{1+e^{2\kappa L}} \right)= ia_0 \frac{e^{2\kappa L} - e^{-2\kappa L}}{(1+e^{2\kappa L})(1+e^{-2\kappa L})}=ia_0 \frac{ \sinh 2\kappa L}{2 \cosh^2 \kappa L} = ia_0 \frac{\sinh \kappa L}{\cosh \kappa L} \]Finally,
\[R = \frac{|b(0)|^2}{|a_0|^2}=\tanh^2 \kappa L\]

A4  ^?? Does the energy loss occur due to the interaction of light with Bragg grating?

Let's find the value of the transmission coefficient $T=|a(L)|^2/|a_0|^2$:
\[a(L)=\frac{a_0 e^{\kappa L}}{1+e^{2 \kappa L}} + \frac{a_0 e^{-\kappa L}}{1+e^{-2 \kappa L}} =a_0 \frac{1}{\cosh \kappa L} \quad \Rightarrow \quad T = \frac{1}{\cosh^2 \kappa L}\]Now we can calculate $R+T$:
\[ R+T=\tanh^2 \kappa L + \frac{1}{\cosh^2 \kappa L} = \frac{\sinh^2 \kappa L+ 1}{\cosh^2 \kappa L} = 1.\]In other words, all incident power is transmitted or reflected and there is no loss.