Finding unknown load using a smith chart
I think the mistake is in how you are using the equation for \$Z_{in}\$. The second time you are using the equation, you forgot to multiply by \$Z_o\$. The equation for this is 

$$Z_{in}=Z_o\dfrac{Z_L+jZ_o\tan{\beta l}}{Z_o+jZ_L\tan{\beta l}}$$ And from what I see, your are missing the \$Z_o\$ factor in part b.

I get, \$Z_L\approx 161-j12\$. 

This can be normalized to the \$Z_o\$ (\$100\Omega\$) impedance in the Smith chart to be \$z_L=1.61-j0.12\$

If omit the \$Z_o\$, I get the same result you've obtained analytically for \$Z_L\$.

Hopefully that matches what you get in the smith chart.

ADD:

As per your question below. In part A, you know the load impedance is an open circuit, that is the far right in the Smith chart. They tell you, the input impedance is \$-j1.5\$. From you load location (at .25\$\lambda\$) to the input impedance (at \$.344\lambda\$) there is a delta of .344 - .25 =0.094\$\lambda\$. That is, you need to move from your load to the location where the measure the input impedance that much (0.094\$\lambda\$) in order to 'see' the given input impedance. So from your load to the input impedance you move toward the source. And the 0.094\$\lambda\$ is the length of the line between the location where they measure the input impedance to the load.

Once they place the load in part B, they give you another the input impedance. You already know the length of the line. So from the new input impedance, you have to move toward the load 0.094\$\lambda\$ in order to find out what the actual load impedance is. That is a CCW rotation. 

From \$Z_{in}\$ to \$Z_L\$, you move CCW and from \$Z_L\$ to \$Z_{in}\$ you move CW.

If you rotate CCW the length found in part A, from the \$z_{in}=1-j0.5\$ given in part B, you will get the same results analytically and on the Smith chart.