One possible problem with your solution is that it contains \(\sqrt t\) in the argument of cosine, when it should be a linear term. Aside from that, the best way to track down a mistake is to start from the beginning:
The mass's position function \(s(t)\) satisfies the second-order ODE
\(4x''+4x'+401x=0\)
(assuming there are no other external forces acting on the mass). The characteristic equation for this ODE is
\(4r^2+4r+401=0\implies r=-\dfrac12\pm10i\)
which means the general solution to this ODE is
\(x(t)=\bigg(C_1\cos(10t)+C_2\sin(10t)\bigg)e^{-t/2}\)
The angle difference identity for cosine allows you to condense the trigonometric part of the solution to
\(C_1\cos(10t)+C_2\sin(10t)=R\cos(10t-\delta)\)
where \(C_1=R\cos\delta\) and \(C_2=R\sin\delta\), leaving you with
\(x(t)=R\cos(10t-\delta)e^{-t/2}\)
These unknown constants can be found explicitly, as \(R=\sqrt{{C_1}^2+{C_2}^2}\) and \(\delta=\arctan\dfrac{c_2}{c_1}\).
Given that \(x(0)=1\) and \(v(0)=x'(0)=7\), and the solution's first derivative is
\(x'(t)=\dfrac12\bigg((20C_2-C_1)\cos(10t)-(20C_1+C_2)\sin(10t)\bigg)e^{-t/2}\)
you have the following system of equations needed to find \(C_1,C_2\), and from there the corresponding values of \(R\) and \(\delta\).
\(\begin{cases}C_1=1\\\\-\dfrac12C_1+10C_2=1\end{cases}\implies C_1=1,C_2=\dfrac3{20}\implies R=\dfrac{\sqrt{409}}{20},\delta=\arctan\dfrac3{20}\)
So the particular solution is
\(x(t)=\dfrac{\sqrt{409}}{20}\cos\left(10t-\arctan\dfrac3{20}\right)e^{-t/2}\)
\(x(t)\approx1.0112\cos(10t-0.1489)e^{-0.5t}\)
In terms of what you should submit, you would use
\(C_1=1.0112\)
\(\omega_1=10\)
\(\alpha_1=0.1489\)
\(p=0.5\)
or rely on the exact forms in case rounded answers are not accepted.
