Joint $L_1$. Members $L_1\!-\!L_2$ (slope $\downarrow$ $3\!:\!4$) and $L_1\!-\!U_1$ (slope $\uparrow$ $3\!:\!4$). $\sum F_x\Rightarrow F_{L_1U_1}=-F_{L_1L_2}$; $\sum F_y:\;60-\tfrac{6}{5}F_{L_1L_2}=0\Rightarrow F_{L_1L_2}=\boxed{50\text{ kN (T)}}$.
Joint $L_4$. Members $L_3\!-\!L_4$, $L_4\!-\!L_5$ ($4\!:\!3$) and vertical $U_3\!-\!L_4$, under the $72\text{ kN}$ load. Solving the joint gives $F_{U_3L_4}=\boxed{30\text{ kN (T)}}$ and $F_{L_3L_4}=\boxed{56\text{ kN (T)}}$.
4(b)
Given. Top chord $U_1\dots U_6$ at $y=6$ (spacing $4\text{ m}$); pinned supports at $L_1(0,0)$ and $L_5(20,3)$; interior lower nodes $L_2(4,3),L_3(12,3),L_4(16,3)$; three $30\text{ kN}$ loads at $U_3,U_4,U_5$.
Find. $U_3\!-\!L_3$, $L_4\!-\!U_6$ and $L_3\!-\!L_4$ (tension/compression).
Note on this truss. Nodes $L_2$ and $L_3$ are joined only through the apex $U_3$ (there is no $L_2\!-\!L_3$ chord), so both feet are pinned; the second pin removes the otherwise-present internal mechanism, and $m+r-2n=18+4-22=0$ — the truss is stable and determinate (the equilibrium matrix has full rank).
4(b) Both feet pinned; requested members in red.
Reactions. Global equilibrium with two pins gives $V_{L_1}=V_{L_5}=45\text{ kN}$ and $H_{L_1}=+60\text{ kN}$, $H_{L_5}=-60\text{ kN}$ (the horizontal thrust is demanded by the two-triangle geometry). Check $\sum M_{L_1}=-1080+45(20)+(-60)(-3)=0$.
Member forces (whole-structure joint solution, tension +): $F_{U_3L_3}=\boxed{25\text{ kN (T)}}$, $F_{L_4U_6}=\boxed{75\text{ kN (T)}}$, and $F_{L_3L_4}=\boxed{0}$ (zero-force member for this loading).
