When calculating the energy released by the $\alpha$ decay of ${}^{238}\text{U}$ you cannot ignore the binding energy of the alpha particle or you will get a spectacularly wrong result. The calculation for the energy released is:

Then the energy released is the binding energy of the products minus the binding energy of the ${}^{238}\text{U}$:

$$ \Delta E = 1777.670 + 28.296 - 1801.696 = 4.270~\text{MeV} $$

Which reassuringly agrees with the value given by Wikipedia. If you ignored the binding energy of the helium nucleus your result would be wrong by $28.296$ MeV! If you'd like to give a link to the page where you saw the He binding energy apparently being ignored we can have a look at it and see what's wrong.

You ask about the hydrogen fusion process. If we take the commercially significant process $D + T \to He + n$ then the energies for this are:

And again the energy released is the BE of the products minus the BE of the two initial nuclei so it's:

$$ \Delta E = 28.296 - (8.482 + 2.225) = 17.589~\text{MeV} $$

So the fusion releases more energy than the fission.

In a comment you ask about the reaction ${}^2D + {}^2D \to {}^3He + n$ and for this the energy is:

$$ \Delta E = 7.718 - (2.225 + 2.225) = 3.268~\text{MeV} $$

and this is indeed less than the energy released in ${}^{238}U$ fission. However I don't think there is a way to see this just by looking at the binding energy per nucleon graph. You need to do the calculation. The trouble is that for the calculation we need the total energy, and that's the BE per nucleon multiplied by the number of nucleons.

When calculating the energy released by the $\alpha$ decay of ${}^{238}\text{U}$ you cannot ignore the binding energy of the alpha particle or you will get a spectacularly wrong result.

You can finding binding energies on this web site. Using this data the calculation for the energy released in ${}^{238}U$ $\alpha$ decay is:

Then the energy released is the binding energy of the products minus the binding energy of the ${}^{238}\text{U}$:

$$ \Delta E = 1777.670 + 28.296 - 1801.696 = 4.270~\text{MeV} $$

Which reassuringly agrees with the value given by Wikipedia. If you ignored the binding energy of the helium nucleus your result would be wrong by $28.296$ MeV! If you'd like to give a link to the page where you saw the He binding energy apparently being ignored we can have a look at it and see what's wrong.

You ask about the hydrogen fusion process. If we take the commercially significant process $D + T \to He + n$ then the energies for this are:

And again the energy released is the BE of the products minus the BE of the two initial nuclei so it's:

$$ \Delta E = 28.296 - (8.482 + 2.225) = 17.589~\text{MeV} $$

So the fusion releases more energy than the fission.

In a comment you ask about the reaction ${}^2D + {}^2D \to {}^3He + n$ and for this the energy is:

$$ \Delta E = 7.718 - (2.225 + 2.225) = 3.268~\text{MeV} $$

and this is indeed less than the energy released in ${}^{238}U$ fission. However I don't think there is a way to see this just by looking at the binding energy per nucleon graph. You need to do the calculation. The trouble is that for the calculation we need the total energy, and that's the BE per nucleon multiplied by the number of nucleons.

When calculating the energy released by the $\alpha$ decay of ${}^{238}\text{U}$ you cannot ignore the binding energy of the alpha particle or you will get a spectacularly wrong result. The calculation for the energy released is:

Then the energy released is the binding energy of the products minus the binding energy of the ${}^{238}\text{U}$:

$$ \Delta E = 1777.670 + 28.296 - 1801.696 = 4.270~\text{MeV} $$

Which reassuringly agrees with the value given by Wikipedia. If you ignored the binding energy of the helium nucleus your result would be wrong by $28.296$ MeV! If you'd like to give a link to the page where you saw the He binding energy apparently being ignored we can have a look at it and see what's wrong.

You ask about the hydrogen fusion process. If we take the commercially significant process $D + T \to He + n$ then the energies for this are:

And again the energy released is the BE of the products minus the BE of the two initial nuclei so it's:

$$ \Delta E = 28.296 - (8.482 + 2.225) = 17.589~\text{MeV} $$

So the fusion releases more energy than the fission.

When calculating the energy released by the $\alpha$ decay of ${}^{238}\text{U}$ you cannot ignore the binding energy of the alpha particle or you will get a spectacularly wrong result. The calculation for the energy released is:

Then the energy released is the binding energy of the products minus the binding energy of the ${}^{238}\text{U}$:

$$ \Delta E = 1777.670 + 28.296 - 1801.696 = 4.270~\text{MeV} $$

Which reassuringly agrees with the value given by Wikipedia. If you ignored the binding energy of the helium nucleus your result would be wrong by $28.296$ MeV! If you'd like to give a link to the page where you saw the He binding energy apparently being ignored we can have a look at it and see what's wrong.

You ask about the hydrogen fusion process. If we take the commercially significant process $D + T \to He + n$ then the energies for this are:

And again the energy released is the BE of the products minus the BE of the two initial nuclei so it's:

$$ \Delta E = 28.296 - (8.482 + 2.225) = 17.589~\text{MeV} $$

So the fusion releases more energy than the fission.

In a comment you ask about the reaction ${}^2D + {}^2D \to {}^3He + n$ and for this the energy is:

$$ \Delta E = 7.718 - (2.225 + 2.225) = 3.268~\text{MeV} $$

and this is indeed less than the energy released in ${}^{238}U$ fission. However I don't think there is a way to see this just by looking at the binding energy per nucleon graph. You need to do the calculation. The trouble is that for the calculation we need the total energy, and that's the BE per nucleon multiplied by the number of nucleons.