For Problem U-2: $\tau _ A($v$_ 1) = A$   v$_ 1 = \left[\begin{array}{rr}4&1\ 1&4\end{array}\right]\left[\begin{array}{r}1\ 1\end{array}\right] = \left[\begin{array}{r}5\ 5\end{array}\right]$, so $\left[\begin{array}{r}1\ 1\end{array}\right]$ has been multiplied by 5. $\tau _ A($v$_ 2) = A$   v$_ 2 = \left[\begin{array}{rr}4&1\ 1&4\end{array}\right]\left[\begin{array}{r}-1\ 1\end{array}\right] = \left[\begin{array}{r}3\ 3\end{array}\right]$, so $\left[\begin{array}{r}-1\ 1\end{array}\right]$ has been multiplied by 3.

For Problem U-3: Just use the definition of the new matrix, carefully. Take v$_ 1$, transform it, and write the result using the basis v$_ 1,$   v$_ 2$, to get the first column of the answer: $\tau _ A($v$_ 1) = \left[\begin{array}{r}5\ 5\end{array}\right] = 5$   v$_ 1 = 5$   v$_ 1 + 0$   v$_ 2$, so the first column is $\left[\begin{array}{r}5\ 0\end{array}\right]$. Similarly, $\tau _ A($v$_ 2) = \dots = 3$   v$_ 2 = 0$   v$_ 1 + 3$   v$_ 2$, so the second column of the answer is $\left[\begin{array}{r}0\ 3\end{array}\right]$. Together they give $\left[\begin{array}{rr}5&0\ 0&3\end{array}\right]$.