Answer: \(\angle BAC=1.24\) rad; perimeter \(\approx15.3\) cm; area \(\approx9.60\text{ cm}^2\).

The arc \(BEC\) has radius \(5\) cm and length \(6.2\) cm. Since arc length is \(r\theta\),

\(5\theta=6.2.\)

So

\(\theta=1.24\) radians.

Thus \(\angle BAC=1.24\) radians.

Since \(D\) is the midpoint of \(BC\), \(BD\) is half of the chord \(BC\). In the right triangle formed by dropping the perpendicular from \(A\) to \(BC\),

\(BD=5\sin\left(\dfrac{1.24}{2}\right)=5\sin0.62=2.91\ldots.\)

The arc \(BFC\) is a semicircle with radius \(BD\), so its length is

\(\pi(2.91\ldots)=9.13\ldots.\)

The perimeter of the shaded region is the sum of the two arcs:

\(6.2+9.13\ldots=15.3\ldots.\)

So the perimeter is

\(\boxed{15.3\text{ cm}}\) to 3 significant figures.

The area of the semicircle is

\(\dfrac12\pi(2.91\ldots)^2=13.3\ldots.\)

The area of sector \(ABC\) is

\(\dfrac12(5)^2(1.24)=15.5.\)

The area of triangle \(ABC\) is

\(\dfrac12(5)^2\sin1.24=11.8\ldots.\)

The circular segment between chord \(BC\) and arc \(BEC\) has area

\(15.5-11.8\ldots=3.73\ldots.\)

Therefore the shaded area is

\(13.3\ldots-3.73\ldots=9.60\ldots.\)

Hence the area is

\(\boxed{9.60\text{ cm}^2}\) to 3 significant figures.