The correct answer is $\boxed{{2 \over 7},0}$.
The Laplace transform of a function $f(t)$ is defined as
$$L\left[ {f\left( t \right)} \right] = \int_0^\infty {f\left( t \right)e^{-st}dt}$$
If $F(s)$ is the Laplace transform of $f(t)$, then the initial and final values of $f(t)$ are given by
$$f(0) = {1 \over s}F\left( s \right)$$
$$f\left( \infty \right) = {1 \over s} lim_{s \to \infty} F\left( s \right)$$
In this case, we have
$$F\left( s \right) = {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}}$$
Therefore,
$$f(0) = {1 \over s}F\left( s \right) = {1 \over s} {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}} = {2 \over 7}$$
and
$$f\left( \infty \right) = {1 \over s} lim_{s \to \infty} F\left( s \right) = {1 \over s} lim_{s \to \infty} {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}} = 0$$
Therefore, the initial and final values of $f(t)$ are respectively ${2 \over 7}$ and $0$.