= Solution
Choose $\boxed{v_4=e^t(1-t)}$, and denote the new ordered <basis> by $\mathcal C=(b_1,b_2,b_3,b_4)$. The <change of basis> <matrix>, with columns expressed in $\mathcal B$, is
$$
P=\begin{pmatrix}
1&1&0&0\\
-1&1&0&0\\
0&0&1&1\\
0&0&1&-1
\end{pmatrix},\qquad\det P=-4\ne0.
$$
This nonzero <determinant> proves that the proposed fourth function completes a <basis>. Express the derivatives in that <basis>:
$$
\begin{aligned}
F(b_1)&=-1=-\tfrac12b_1-\tfrac12b_2,\\
F(b_2)&=1=\tfrac12b_1+\tfrac12b_2,\\
F(b_3)&=e^t(2+t)=\tfrac32b_3+\tfrac12b_4,\\
F(b_4)&=-te^t=-\tfrac12b_3+\tfrac12b_4.
\end{aligned}
$$
Reading off columns therefore gives
$$
\boxed{[F]_{\mathcal C}=P^{-1}MP=
\begin{pmatrix}
-\tfrac12&\tfrac12&0&0\\
-\tfrac12&\tfrac12&0&0\\
0&0&\tfrac32&-\tfrac12\\
0&0&\tfrac12&\tfrac12
\end{pmatrix}.}
$$
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