Answer
See the detailed answer below.
Work Step by Step
a) The net effect is similar to the proton-proton chain since there is no carbon consumed in the carbon cycle, as we see below.
$$\rm \color{blue}{\bf^{12}_6C}+\;^1_1H\rightarrow \;\;\color{red}{^{13}_7N}+\gamma\tag 1$$
where $\rm ^{13}_7N$ decays to
$$\rm ^{13}_7N\rightarrow \;\;\color{red}{^{13}_6C}+e^++\nu\tag2$$
and hence,
$$\rm ^{13}_6C+\;^1_1H\rightarrow \;\;\color{red}{^{14}_7N}+\gamma\tag3$$
and then
$$\rm ^{14}_7N+\;^1_1H\rightarrow \;\;\color{red}{^{15}_8O}+\gamma\tag4$$
where $\rm {^{15}_8O}$ decays to
$$\rm ^{15}_8O\rightarrow \;\;\color{red}{^{15}_7N}+e^++\nu\tag5$$
and
$$\rm ^{15}_7N+\;^1_1H\rightarrow \;\;\color{blue}{\bf^{12}_6C}+ \;^4_2He\tag6$$
As we see, $\color{blue}{\bf^{12}_6C}$ finally comes out as a reaction result after a sequence of reactions and decays.
And to make sure, let's add all these reactions together and see the resultant.
$$\begin{gather*}\rm \color{blue}{\bf^{12}_6C}+\;^1_1H + \; ^{13}_7N+\;^{13}_6C+\;^1_1H+ \;^{14}_7N+\\
\rm\;^1_1H+\; ^{15}_8O+\;^{15}_7N+\;^1_1H \rightarrow\\\\
\rm {^{13}_7N}+\gamma+{^{13}_6C}+e^++\nu +{^{14}_7N}+\gamma+\\
\rm {^{15}_8O}+\gamma+{^{15}_7N}+e^++\nu +\color{blue}{\bf^{12}_6C}+ \;^4_2He \end{gather*}$$
Therefore, the net reaction is given by
$$\boxed{\rm\;4\left(^1_1H\right)\rightarrow\;\;^4_2He+2e^++3\gamma+2\nu}$$
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b) To find the energy produced we need to balance the previous boxed formula in number of electrons.
Hence, it will be as follows;
$$ \rm\;4\left(^1_1H\right)\rightarrow\;\;^4_2He+2e^++2e^-$$
Thus, the produced energy is given by
$$ Q=\;4Q_\left(\rm ^1_1H\right)-Q_{\rm^4_2He}-4Q_e$$
$$ Q=\left[4m_\left(\rm ^1_1H\right)-m_{\rm^4_2He}-4m_e\right]c^2$$
Plugging the known form appenix B;
$$ Q=\left[4(1.007825)-4.002603-4\left(\dfrac{9.11\times 10^{-31}}{1.66\times 10^{-27}}\right)\right]931.5$$
$$ Q=\color{red}{\bf 24.69}\;\rm MeV$$
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c)
$\Rightarrow$ For the first reaction above, (1), we do not need to rewrite it, and hence the energy released is given by
$$ Q=\left[m_{\rm^{12}_6C}-m_{\rm^1_1H}-m_{\rm^{13}_7N}\right]c^2$$
$$ Q_1=\left[12.000000+1.007825- 13.005739\right]931.5$$
$$Q_1=\color{red}{\bf 1.94311}\;\rm MeV$$
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$\Rightarrow$ For the second reaction above, (2), we need to balance its electrons in both sides; $\rm ^{13}_7N\rightarrow \;\; {^{13}_6C}+e^++e^-+\nu$, and hence the energy released is given by
$$ Q =\left[13.005739-13.003355- 2\left(\dfrac{9.11\times 10^{-31}}{1.66\times 10^{-27}}\right)\right]931.5=\bf 1.19829\;\rm
MeV$$
Recall that, in this reaction, there is an electron–positron annihilation, so that
$$Q_2 =1.19829+1.02=\color{red}{\bf 2.21829 }\;\rm MeV$$
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$\Rightarrow$ For the third reaction above, (3), we do not need to rewrite it, and hence the energy released is given by
$$ Q_3=\left[13.003355+1.007825- 14.003074\right]931.5$$
$$Q_3=\color{red}{\bf 7.55074}\;\rm MeV$$
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$\Rightarrow$ For the fourth reaction above, (4), we do not need to rewrite it, and hence the energy released is given by
$$ Q_4=\left[14.003074+1.007825- 15.003066\right]931.5$$
$$Q_4=\color{red}{\bf 7.29644}\;\rm MeV$$
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$\Rightarrow$ For the fifth reaction above, (5), wwe need to balance its electrons in both sides; $\rm ^{15}_8O\rightarrow \;\;\color{red}{^{15}_7N}+e^++e^-+\nu$, and hence the energy released is given by
$$ Q =\left[15.003066-15.000109 - 2\left(\dfrac{9.11\times 10^{-31}}{1.66\times 10^{-27}}\right)\right]931.5=\bf 1.73204\;\rm
MeV$$
Recall that, in this reaction, there is an electron–positron annihilation, so that
$$Q_5 =1.73204+1.02=\color{red}{\bf 2.75204}\;\rm MeV$$
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$\Rightarrow$ For the sixth reaction above, (6), we do not need to rewrite it, and hence the energy released is given by
$$ Q_6=\left[15.000109 +1.007825- 12.000000-4.002603\right]931.5$$
$$Q_6=\color{red}{\bf 4.96583}\;\rm MeV$$
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d) We know that this reaction of carbon cycle needs more temperature than the proton-proton chain because the atoms need to have more kinetic energy to conquer the repulsion force between the nuclei.
