If you're going to do a lot of work in 4 dimensions, it might be worth learning two-component notation. This is used in the text books, e.g., of Srednicki, Buchbinder & Kuzenko or the comprehensive article by Dreiner, Haber and Martin.
Two component notation has the advantage that the 4 dimensional objects decompose into the direct sum (or something similar) of 2 dimensional objects, which have fewer invariants and identities.
I'll use the conventions of B&K (which I think slightly differ from those in the question).
The four component spinors takes the form
$$
\psi_a = \begin{pmatrix}\psi_\alpha \\ \bar\psi^{\dot\alpha}\end{pmatrix} \,, \quad
\bar\psi^a = \begin{pmatrix}\psi^\alpha & \bar\psi_{\dot\alpha}\end{pmatrix}
$$
and the gamma and projection matrices
$$
\gamma_a = \begin{pmatrix} 0 & \sigma_a \\ \tilde\sigma_a & 0 \end{pmatrix}\,,\quad
\gamma_5 = -i \gamma_0\gamma_1\gamma_2\gamma_3
= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \,,\quad
$$
$$
P_L = \frac12 (1 + \gamma_5) \,, \quad P_R = \frac12 (1 - \gamma_5)
$$
the left and right spinors are then
$$
\psi_L = P_L\psi = \begin{pmatrix}\psi \\ 0 \end{pmatrix} \,, \quad
\psi_R = P_R\psi = \begin{pmatrix} 0 \\ \bar\psi \end{pmatrix} \,, \quad
$$
where there should never be any reason to confuse a 2-component and 4-component spinor.
The spinor bilinears are
$$
\bar\theta \theta = \theta^2 + \bar\theta^2 \,,\quad
\bar\theta \gamma_5 \theta = \theta^2 - \bar\theta^2
$$
etc..., where the contraction convention is
$\theta^2 = \theta^\alpha \theta_\alpha$ and
$\bar\theta^2 = \bar\theta_{\dot\alpha} \bar\theta^{\dot\alpha}$.
Then, to check your final identity (for example), the left hand side is
$$ (\bar\theta\gamma_5\theta)(\bar\psi_L\theta)(\bar\theta\lambda)
= (\theta^2-\bar\theta^2)(\bar\psi\bar\theta)(\theta\lambda+\bar\theta\bar\lambda)
= (\theta^2-0)(-\frac12\bar\theta^2)(\bar\psi\bar\lambda)
= -\frac12\theta^2\,\bar\theta^2\,\bar\psi\bar\lambda
$$
where we used the fact that
$\theta^3=\theta_{\alpha_1}\theta_{\alpha_2}\theta_{\alpha_3}=0$
and similarly for $\bar\theta^3=0$.
We also used the important identities
$$ \begin{align}
\theta_\alpha \theta^\beta &= -\frac12\theta^2\delta_\alpha^\beta \,,&
\bar\theta^{\dot\alpha}\bar\theta_{\dot\beta} &=-\frac12\bar\theta^2\delta_{\dot\beta}^{\dot\alpha} \ .
\end{align}$$
The right hand side is
$$ \frac14(\bar\theta\gamma_5\theta)^2(\bar\psi_L\lambda)
= \frac14(\theta^2-\bar\theta^2)(\theta^2-\bar\theta^2)(\bar\psi\bar\lambda)
= -\frac12\theta^2\bar\theta^2\bar\psi\bar\lambda
$$
so the last identity checks out.
The rest of the identities can be similarly checked.
Edit:
Since there seems to be a bit of confusion for the other identities, here's their proofs.
Because I'm lazy, I've suppressed all indices. For example
$\bar\theta\partial\psi
=\bar\theta_{\dot\alpha}\tilde\sigma^{\dot\alpha\alpha}_a\partial^a\psi_\alpha
=-(\partial^a\psi_\alpha)\tilde\sigma^{\dot\alpha\alpha}_a\bar\theta_{\dot\alpha}
=-(\partial_a\psi^\alpha)\sigma_{\alpha\dot\alpha}^a\bar\theta^{\dot\alpha}
=-\psi\overleftarrow{\partial}\bar\theta\ .$
Identity 1:
$$ LHS = (\theta^2-\bar\theta^2)(\bar\psi\bar\theta)(\bar\theta\partial\psi)
= (\theta^2-0)(-\frac12\bar\theta^2)(\bar\psi\partial\psi)
= -\frac12\theta^2\bar\theta^2(\bar\psi\partial\psi)
$$
$$ RHS = \frac14(\theta^2-\bar\theta^2)^2(\bar\psi\partial\psi) = LHS
$$
Identity 2 is basically just the complex conjugate result.
Identity 3: I think you have a mistake in this one (ignoring the obvious problem of the non-matching $\mu$ index). This is because $(\bar{\psi}\gamma ^\mu \psi)=0$ if we insert a $\gamma_5$ so it's like the term on the left it's nonvanishing $(\bar{\psi}\gamma_5\gamma_\mu \psi)=2\psi\sigma_\mu\bar\psi$. So
$$
LHS
= (\theta\sigma_\mu\bar\theta-\bar\theta\tilde\sigma_\mu\theta)(\bar\psi\bar\theta)(\psi\theta)
= 2(\psi\theta)(\theta\sigma_\mu\bar\theta)(\bar\theta\bar\psi)
= \frac12\theta^2\bar\theta^2 (\psi\sigma_\mu\bar\psi)
$$
$$
RHS = \frac{1}{4} (\bar{\theta}\gamma _5 \theta)^2 (\bar{\psi}\gamma_5\gamma_\mu\psi)
= -\theta^2\bar\theta^2(\psi\sigma_\mu\bar\psi)
$$
so it seems to be out by a factor of $-1/2$. (Although, I might have miscounted the factors of 1/2).
