Answer:$(1)L^{-1}\bigg[\dfrac{2s-1}{s^2+4s+29}\bigg]$

$L^{-1}\bigg[\dfrac{2s-1}{s^2+4s+29}\bigg]=L^{-1}\bigg[\dfrac{2s-1}{(s+2)^2+5^2}\bigg]$

=$L^{-1}\bigg[\dfrac{2s-1+4-4}{(s+2)^2+5^2}\bigg] $

=$L^{-1}\bigg[\dfrac{2s+4}{(s+2)^2+5^2}\bigg]-L^{-1}\bigg[\dfrac{5}{(s+2)^2+5^2}\bigg]$

$=L^{-1}\bigg[\dfrac{2(s+2)}{(s+2)^2+5^2}\bigg]-5L^{-1}\bigg[\dfrac{1}{(s+2)^2+5^2}\bigg]$

$=2e^{-2t}L^{-1}\bigg[\dfrac{s}{(s)^2+5^2}\bigg]-5e^{-2t}L^{-1}\bigg[\dfrac{1}{(s)^2+5^2}\bigg] $

=$2e^{-2t}\cos5t-e^{-2t}\sin5t$

$(2)L^{-1}\bigg[\cot^{-1}\bigg(\dfrac{s+3}{2}\bigg)\bigg]$

$L^{-1}\bigg[\cot^{-1}\bigg(\dfrac{s+3}{2}\bigg)\bigg]=\dfrac{-1}{t}L^{-1}\bigg[\dfrac{\mathrm{d} }{\mathrm{d} s}\cot^{-1}\bigg(\dfrac{s+3}{2}\bigg)\bigg] $

=$\dfrac{-1}{t}L^{-1}\Bigg[\dfrac{-1}{1+\Big(\dfrac{s+3}{2}\Big)^{2}}\times\dfrac{1}{2}\Bigg]$

$=\dfrac{2}{t}L^{-1}\Bigg[\dfrac{1}{1^2+(s+3)^{2}}\Bigg]$

$=\dfrac{2}{t}e^{-3t}L^{-1}\Bigg[\dfrac{1}{1^2+(s)^{2}}\Bigg] $

$=\dfrac{2}{t}e^{-3t} \sin t$