We have the following IVP for $w\in C^3(0,0.5)$: $$\begin{cases}w'''(t)+w''(t)=e^t+1 \\ w(0)=1; w'(0)=0; w''(0)=1\end{cases}.$$
The question is to do an appropriate change of variable so that we get 3 first order IVPs which we know how to solve numerically.
My Try: If we use the change of variable $v(t)\to w''(t)$, we get the following first order system: $$\begin{cases}v'(t)+v(t)=e^t+1 \\ v(0)=1\end{cases}.$$
The given system then gives us an approximation of $v$ using rudimentary methods (Runge-Kutta). Now, this is only one 1st order system. The question asks for two more; what are they?
Define $\boldsymbol{y} := (w,w',w'')^{\top}$, then you obtain from the ODE: \begin{equation} \boldsymbol{y}' = \left( \begin{array}{c} w'\\ w''\\ w''' \end{array} \right) = \left( \begin{array}{c} w'\\ w''\\ -w'' + e^t + 1 \end{array} \right) =: \boldsymbol{f}(t,\boldsymbol{y}), \end{equation} and from the given initial conditions: \begin{equation} \boldsymbol{y}(0) = \left( \begin{array}{c} w(0)\\ w'(0)\\ w''(0) \end{array} \right) = \left( \begin{array}{c} 1\\ 0\\ 1 \end{array} \right) =: \boldsymbol{y}_0. \end{equation} You now have an initial-value problem in the standard form: $\boldsymbol{y}' = \boldsymbol{f}(t,\boldsymbol{y})$, $\boldsymbol{y}(0) = \boldsymbol{y}_0$, which you can solve numerically using a Runge-Kutta method.