= Solution
Put $a=\pi/2$. Substituting $x=-a$ and $x=a$ into part (ii), using $\cos(\pi/4)=\sin(\pi/4)=1/\sqrt2$, gives
$$
-1+\frac{e^a}{\sqrt2}(A-B)=e^a,
\qquad
1+\frac{e^{-a}}{\sqrt2}(A+B)=e^{-a}.
$$
Thus $A-B=\sqrt2(1+e^{-a})$ and $A+B=\sqrt2(1-e^a)$. Solving these two <linear equations> gives
$$
A=\sqrt2(1-\sinh a),\qquad B=-\sqrt2\cosh a.
$$
Consequently the solution of the <boundary value problem> is
$$
\boxed{y(x)=\sin x+\sqrt2 e^{-x}\left[\left(1-\sinh\frac\pi2\right)\cos\frac x2-\cosh\frac\pi2\sin\frac x2\right].}
$$
The two independent <boundary conditions> determine $A,B$ uniquely.
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