Assuming you have got the fractions correctly, the third fraction could be done as follows:
$$\frac{s}{(s+1)^2+4}=\frac{s\color{red}{+1}\color{blue}{-1}}{(s+1)^2+4}=\frac{s\color{red}{+1}}{(s+1)^2+4}+\frac{\color{blue}{-1}}{(s+1)^2+4}\\=\frac{S}{S^2+4}|_{S\to s\color{red}{+1}}\color{blue}{-}\frac{1}2\frac{2}{S^2+4}|_{S\to s+1}$$
$$=e^{-t}\cos 2t\color{blue}{-}\frac{1}2e^{-t}\sin 2t$$
Note that: $$\mathcal{L}(e^{at}f(t))=F(s)|_{s\to s-1}$$
We are given:
$$y''+2y'+y=0; \;\;\;\;\;y'(0)=2\;\;\;\;\;\mbox{and}\;\;\;\;\;y(1)=2$$
I am going to start you off, but you have to show some work and fill in the details.
Taking the Laplace transform yields:
$$\mathscr{L}(y''+2y'+y) = (s^2 y(s) -s y(0) -y'(0)) +2 (s y(s) - y(0)) + y(s) = 0$$
Substituting the ICs (let $y(0) = a$) yields:
$$(s^2 y(s) -s a -2) +2 (s y(s) - a) + y(s) = 0$$
Solving for $y(s)$ yields:
$$y(s) = \dfrac{as +2a+2}{(s+1)^2}$$
Now, find the inverse Laplace transform:
$$y(t) = e^{-t}(at + a + 2t)$$
Next, substitute in $y(1) = 2$ and solve for $a$ yielding:
$$y(t) = e^{-t}(e t + t + e - 1)$$
Best Answer
Laplace-transforming both sides of $y'' + y = 2t$,
$$s^2 Y (s) - y_0 s - v_0 + Y (s) = \frac{2}{s^2}$$
Hence,
$$Y (s) = y_0 \left(\frac{s}{s^2 + 1}\right) + (v_0 - 2)\left(\frac{1}{s^2 + 1}\right) + \frac{2}{s^2}$$
Taking the inverse Laplace transform,
$$y (t) = y_0 \cos (t) + (v_0 - 2) \sin (t) + 2 t$$
From the conditions $y(\pi /4) = \pi / 2 $ and $y'(\pi /4) = 2- \sqrt{2}$, we get
$$y_0 + v_0 = 2 \qquad \qquad \qquad y_0 - v_0 = 2$$
Thus, $y_0 = 2$ and $v_0 = 0$, and
$$y (t) = 2 \cos (t) - 2 \sin (t) + 2 t$$