Let one sum represent $a+d$ and another represent $b+c$. Translate the statement about the two pair means into a relationship between those sums.

Once you know $a+d$, use the range $d-a=64$. Adding those two equations isolates the larger endpoint.

Algebra is faster here because the problem gives relationships among sums of values. Desmos can verify the two group sums.

Let $x$ represent $b+c$ and let $y$ represent $a+d$. Enter $y=137-x$ and $y=x+8$. Click their intersection; its $y$-coordinate is the endpoint sum $a+d$.

In a new expression line, enter $((145/2)+64)/2$. This combines the endpoint sum with the range to calculate the larger endpoint.

Because there are $9$ data values and their mean is $34$, their total is $9(34)=306$.

The five known values have total

Therefore,

Let $E=a+d$ and $I=b+c$. Then

The mean of $a$ and $d$ is $4$ greater than the mean of $b$ and $c$, so

Thus $E=I+8$. Combining this with $E+I=137$ gives

Since the values are in ascending order, the range is $d-a=64$. Add this equation to $d+a=\frac{145}{2}$:

Therefore,