One approach is to assume that you can split $u(x,t)=X(x)\cdot T(t)$. I don't know the english expression, but it should be something like "separation approach". Using this for the equation $u_{tt}=u_{xx}$, we get \begin{align} \frac{d^2T}{dt^2} \cdot X = T \cdot \frac{d^2X}{dx^2} \\ \Leftrightarrow \frac{d^2T}{dt^2} \cdot T^{-1} = X^{-1} \frac{d^2X}{dx^2}X \equiv -\lambda^2 \end{align} Where $\lambda \in C$ is an arbitrary constant. The reason, why the expression are constant, is that the left side only depends on $t$ while the right side only depends on $x$. If now the i.e. the left side would not be constant, we couldn't reproduce that change on the right side as it is only depend on $x$. The reason why I choose $\lambda^2$ and not $-\lambda$ or $+\lambda^2$ becomes clear in the next step. From the above it follows \begin{align} &\frac{d^2T}{dt^2} = -\lambda^2 T \Rightarrow T(t) = a \sin(\lambda t) + b \cos(\lambda t) \\& \frac{d^2X}{dt^2} = -\lambda^2 X \Rightarrow X(x)=c \sin(\lambda x) + d \cos(\lambda x) \end{align} If we had chosen $\lambda^2$ instead of $-\lambda^2$ we would have got another solution to the two ODEs, i.e. the $\exp()$. But as the initial condition suggest trigonometric functions, $\sin()$ and $\cos()$ fit better. Okay. So we have $u(x,t) =X(x)T(t)$ Now we check for initial condition. \begin{align} u(x,0) = X(x)\cdot T(0) = \cos(\frac{\pi x}{2}) \end{align} This suggests $c=0,\lambda=\frac{\pi}{2}$ and with $T(0) = b$ it follows $d\cdot b=1$ and $X(x) =d\cdot \cos(\frac{\pi x}{2})$.
Next initial condition is $u_t(x,0)=0$. \begin{align} u_t(x,0) &= T'(t)X(0) =(\frac{\pi}{2} a \cos(\frac{\pi}{2} 0) -\frac{\pi}{2} b \sin(\frac{\pi}{2} 0)) \cdot d\cos(\frac{\pi x}{2}) = \\ &=(\frac{\pi}{2} a)\cdot d\cos(\frac{\pi x}{2})=0 \end{align} As $d\neq0$, it follows $a=0$ and therefore $T(t) = b\cos(\frac{\pi t}{2})$. Combining this, we have \begin{align} u(x,t) = b\cos(\frac{\pi t}{2})\cdot d \cos(\frac{\pi x}{2})=e\cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2}) \end{align} With $e=b\cdot d$. But as $b\cdot d = 1$ as stated above, the solution becomes \begin{align} u(x,t) = \cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2}) \end{align} Now we still have to check the last initial condition $u_x(0,t)=0$. \begin{align} u_x(0,t) = X'(x)T(0) = \frac{\pi}{2} \sin(0)T(0) = 0 \end{align} So this one also holds. Therefore we have the solution to the PDE as $u(x,t) = \cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2})$.
Note: In my opinion we could have gone a lot faster. Why? Except for the first initial condition the problem looks quite symmetric. So thought that maybe also $T(t) = \cos(\frac{\pi t}{2})$. And if you check this with the first initial condition it holds.