= Solution
For the first <Butcher tableau>, $b=(1/2,1/2)^T$ and the off-diagonal entries sum to $1/2$, while each diagonal entry is $1/4$. Thus
$$
BA+A^TB-bb^T=\begin{pmatrix}0&0\\0&0\end{pmatrix}.
$$
The positive weights and zero <matrix> prove that \b[method 1 is <algebraically stable>]. It is the two-stage <Gauss collocation method>.
The second method is explicit and has $b=(1/4,3/8,3/8)^T$. Its first diagonal algebraic-stability entry is
$$
m_{11}=2b_1a_{11}-b_1^2=-\frac1{16}<0.
$$
A <positive semidefinite> <matrix> cannot have a negative diagonal <quadratic form>, so \b[method 2 is not <algebraically stable>]. The negative value supplies a direct test; its order is irrelevant to this conclusion.
For the third method, $A=\begin{pmatrix}1/4&-1/4\\1/4&5/12\end{pmatrix}$ and $b=(1/4,3/4)^T$. Direct computation gives
$$
M=\frac1{16}\begin{pmatrix}1&-1\\-1&1\end{pmatrix},\qquad
v^TMv=\frac1{16}(v_1-v_2)^2\geq0.
$$
Both weights are positive, so \b[method 3 is <algebraically stable>].
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