In 4.3(a) start by asking if the \$-3\:\text{V}\$ is more negative than either or both of the sources at the diode anodes. The answer is yes in this case. So you know that at least one of the two diodes is on. Next, if either \$D_1\$ or \$D_2\$ is on then \$V\$ is either \$+2\:\text{V}\$ or else \$+1\:\text{V}\$. If \$V=+1\:\text{V}\$ then this would violate the ideal case for \$D_2\$, which would now be forward-biased by \$1\:\text{V}\$. So \$V=+2\:\text{V}\$, diode \$D_2\$ is on and at this point you know, by definition, that \$D_1\$ must be off. Done.
In 4.3(b) start by asking if the \$+3\:\text{V}\$ is more positive than either or both of the sources at the diode cathods. The answer is yes in this case, too. So you know that at least one of the two diodes is on. Next, if either \$D_1\$ or \$D_2\$ is on then \$V\$ is either \$+2\:\text{V}\$ or else \$+1\:\text{V}\$. (Same question as before.) If \$V=+2\:\text{V}\$ then this would violate the ideal case for \$D_1\$, which would now be forward-biased by \$1\:\text{V}\$. So \$V=+1\:\text{V}\$, diode \$D_1\$ is on and at this point you know, by definition, that \$D_2\$ must be off. Done, again.