Given Lagrangian densities, for example:
$ L = \partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi^2 +\lambda \phi(x)$,
the Euler-Lagrange equation yields
$\partial^2 \phi + m^2 \phi = \lambda $.
Please, very explicitly describe, where does the first time come from ($\partial^2 \phi$)?
Also, how does deriving the equations of motion for $A_\mu$ work for a Lagrangian density such as
$L = -\frac{1}{4}F_{\mu\nu} F^{\mu\nu}$,
where
$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$.
Note that you dropped a $\frac{1}{2}$ in a term in your first Lagrangian density (hereby shorted to Lagrangian). It should be $$\mathcal{L}=\frac{1}{2}\partial_\mu\partial^\mu\phi-\frac{1}{2}m^2\phi^2+\lambda\phi$$
The Euler-Lagrange equations are, for a scalar field, as follows $$\frac{\partial\mathcal{L}}{\partial\phi}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)$$ Now, whenever we directly plug in the Lagrangian, we must have indices in the Lagrangian differ from the ones in the equation, and that the index placement is the same. So, your first Lagrangian becomes, assuming we are working in Minkowski space, $$\mathcal{L}=\frac{1}{2}\eta^{\alpha\beta}\partial_\alpha\phi\partial_\beta\phi-\frac{1}{2}m^2\phi^2+\lambda\phi$$ Plugging in, we have $$-m^2\phi+\lambda=\frac{1}{2}\partial_\mu\left(\eta^{\alpha\beta}\frac{\partial(\partial_\alpha\phi\partial_\beta\phi)}{\partial(\partial_\mu\phi)}\right)$$ Using the product rule and the rule that $\frac{\partial A_\mu}{\partial A_\nu}=\delta^\nu_\mu$ $$ \begin{align} -m^2\phi+\lambda&=\frac{1}{2}\eta^{\alpha\beta}\partial_\mu\left(\partial_\beta\phi\frac{\partial(\partial_\alpha\phi)}{\partial(\partial_\mu\phi)}+\partial_\alpha\phi\frac{\partial(\partial_\beta\phi)}{\partial(\partial_\mu\phi)}\right) \\ &=\frac{1}{2}\eta^{\alpha\beta}\partial_\mu\left(\partial_\beta\phi\delta^\mu_\alpha+\partial_\alpha\phi\delta^\mu_\beta\right)\\ &=\frac{1}{2}\left(\partial_\mu\partial^\alpha\phi\delta^\mu_\alpha+\partial_\mu\partial^\beta\phi\delta^\mu_\beta\right)\\ &=\frac{1}{2}\left(\partial_\mu\partial^\mu\phi+\partial_\mu\partial^\mu\phi\right)\\ &=\partial^\mu\partial_\mu\phi = \partial^2\phi \end{align} $$ So, we have $$\partial^2\phi+m^2\phi=\lambda$$
As for the second part of your question, it is rather similar. The Euler-Lagrange equations now read $$\frac{\partial\mathcal{L}}{\partial A_\nu}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\right)$$ and the Lagrangian reads, abiding by our two rules, $$\begin{align}\mathcal{L}&=-\frac{1}{4}\eta^{\alpha\rho}\eta^{\beta\sigma}F_{\alpha\beta}F_{\rho\sigma}\\ &=-\frac{1}{4}\eta^{\alpha\rho}\eta^{\beta\sigma}(\partial_\alpha A_\beta -\partial_\beta A_\alpha)(\partial_\rho A_\sigma -\partial_\sigma A_\rho) \end{align}$$ Plugging in, we get $$ \begin{align} 0 &= -\frac{1}{4}\eta^{\alpha\rho}\eta^{\beta\sigma}\partial_\mu\left(\frac{\partial(\partial_\alpha A_\beta -\partial_\beta A_\alpha)(\partial_\rho A_\sigma -\partial_\sigma A_\rho)}{\partial(\partial_\mu A_\nu)}\right)\\ &=-\frac{1}{4}\eta^{\alpha\rho}\eta^{\beta\sigma}\partial_\mu\left((\partial_\rho A_\sigma -\partial_\sigma A_\rho)\left(\frac{\partial(\partial_\alpha A_\beta)}{\partial(\partial_\mu A_\nu)}-\frac{\partial(\partial_\beta A_\alpha)}{\partial(\partial_\mu A_\nu)}\right)+(\partial_\alpha A_\beta -\partial_\beta A_\alpha)\left(\frac{\partial(\partial_\rho A_\sigma)}{\partial(\partial_\mu A_\nu)}-\frac{\partial(\partial_\sigma A_\rho)}{\partial(\partial_\mu A_\nu)}\right)\right)\\ &=-\frac{1}{4}\eta^{\alpha\rho}\eta^{\beta\sigma}\partial_\mu\left((\partial_\rho A_\sigma -\partial_\sigma A_\rho)\left(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha\right)+(\partial_\alpha A_\beta -\partial_\beta A_\alpha)\left(\delta^\mu_\rho\delta^\nu_\sigma-\delta^\mu_\sigma\delta^\nu_\rho\right)\right)\\ &=-\frac{1}{4}\partial_\mu\left((\partial_\rho A_\sigma -\partial_\sigma A_\rho)\left(\eta^{\mu\rho}\eta^{\nu\sigma}-\eta^{\nu\rho}\eta^{\mu\sigma}\right)+(\partial_\alpha A_\beta -\partial_\beta A_\alpha)\left(\eta^{\alpha\mu}\eta^{\beta\nu}-\eta^{\alpha\nu}\eta^{\beta\mu}\right)\right)\\ &=-\frac{1}{4}\partial_\mu\left(\partial^\mu A^\nu -\partial^\nu A^\mu-\partial^\nu A^\mu+\partial^\mu A^\nu+\partial^\mu A^\nu -\partial^\nu A^\mu-\partial^\nu A^\mu+\partial^\mu A^\nu\right)\\ &=-\frac{1}{4}\partial_\mu\left(F^{\mu\nu}+F^{\mu\nu}+F^{\mu\nu}+F^{\mu\nu}\right)\\ &=-\partial_\mu F^{\mu\nu} \end{align} $$ So, at the end of all that, we have $$\partial_\mu F^{\mu\nu}=0$$
