(a) Expand the given expression as
[tex]\dfrac{3s-10}{s^2+25}=3\cdot\dfrac s{s^2+25}-2\cdot\dfrac5{s^2+25}[/tex]
You should recognize the Laplace transform of sine and cosine:
[tex]L[\cos(at)]=\dfrac s{s^2+a^2}[/tex]
[tex]L[\sin(at)]=\dfrac a{s^2+a^2}[/tex]
So we have
[tex]L^{-1}\left[\dfrac{3s-10}{s^2+25}\right]=3\cos(5t)-2\sin(5t)[/tex]
(b) Take the Laplace transform of both sides:
[tex]y'(t)+3y(t)=e^{6t}\implies (sY(s)-y(0))+3Y(s)=\dfrac1{s-6}[/tex]
Solve for [tex]Y(s)[/tex]:
[tex](s+3)Y(s)-2=\dfrac1{s-6}\implies Y(s)=\dfrac{2s-11}{(s-6)(s+3)}[/tex]
Decompose the right side into partial fractions:
[tex]\dfrac{2s-11}{(s-6)(s+3)}=\dfrac{\theta_1}{s-6}+\dfrac{\theta_2}{s+3}[/tex]
[tex]2s-11=\theta_1(s+3)+\theta_2(s-6)[/tex]
[tex]2s-11=(\theta_1+\theta_2)s+(3\theta_1-6\theta_2)[/tex]
[tex]\begin{cases}\theta_1+\theta_2=2\\3\theta_1-6\theta_2=-11\end{cases}\implies\theta_1=\dfrac19,\theta_2=\dfrac{17}9[/tex]
So we have
[tex]Y(s)=\dfrac19\cdot\dfrac1{s-6}+\dfrac{17}9\cdot\dfrac1{s+3}[/tex]
and taking the inverse transforms of both sides gives
[tex]y(t)=\dfrac19e^{6t}+\dfrac{17}9e^{-3t}[/tex]
