Answer
See the detailed answer below,
Work Step by Step
In this problem, we will use the formula suggested by the author in all parts. We need to use Appendix B for the atomic mass number and atomic mass.
We can find the excess mass in $\rm u$ and the author asks to find it in
$\rm MeV/c^2$
Recalling that $$\rm 1 \;u= 931.5\;{\rm MeV}/c^2\tag 1$$
a)
$${ \Delta m_{\rm\left[^{4}_2He\right]}}=m_{\rm ^{4}_2He}-A_{\rm ^{4}_2He}=4.002603-4=\color{red}{\bf 0.002603}\;\rm u$$
Using (1);
$${ \Delta m_{\rm\left[^{4}_2He\right]}}=0.002603\times 931.5=\color{blue}{\bf 2.4246945}\;{\rm MeV}/c^2$$
b)
$${ \Delta m_\left[\rm^{12}_6C\right]}=m_{\rm ^{12}_6C}-A_{\rm ^{12}_6C}=12.000000-12=\color{red}{\bf 0}\;\rm u$$
$${ \Delta m_\left[\rm^{12}_6C\right]}=0\times 931.5=\color{blue}{\bf 0}\;{\rm MeV}/c^2$$
c)
$${ \Delta m_\left[\rm ^{86}_{38}Sr\right]}=m_{\rm ^{86}_{38}Sr}-A_{\rm ^{86}_{38}Sr}=85.909261-86=\color{red}{\bf -0.90739}\;\rm u$$
$${ \Delta m_\left[\rm ^{86}_{38}Sr\right]}=-0.90739\times 931.5=\color{blue}{\bf -845.233785}\;{\rm MeV}/c^2$$
d)
$${ \Delta m_\left[\rm ^{235}_{92}U\right]}=m_{\rm ^{235}_{92}U}-A_{\rm ^{235}_{92}U}=235.043930-235=\color{red}{\bf 0.043930}\;\rm u$$
$${ \Delta m_\left[\rm ^{235}_{92}U\right]}=0.043930\times 931.5=\color{blue}{\bf 40.920795}\;{\rm MeV}/c^2$$
e) From Appendix B, we can figure out that the sign of $\Delta$ is positive from $A=1$ to $A=15$ and from $A=218$ to $A=294$.
And the sign of $\Delta$ is negative from $A=16$ to $A=214$.
Therefore,
$$\boxed{\Delta \gt0{\rm \;when}\rightarrow\;\;\;(1\leq A\leq 15), {\;\rm and} \;(218\leq A\leq 294)}$$
$$\boxed{ \Delta \lt0{\rm \;when}\rightarrow\;\;\;(16\leq A\leq 214) }$$