= Solution
Expand the exact solution about $t_{n+1}=t$. The left side is
$$
y(t+h)-y(t)
=hy'+\frac{h^2}{2}y''
+\frac{h^3}{6}y'''
+\frac{h^4}{24}y^{(4)}+O(h^5).
$$
Since $f(t,y(t))=y'(t)$, the right side expands as
$$
hy'
+h^2(1+2\alpha+\beta)y''
+\frac{h^3}{2}(1-\beta)y'''
+\frac{h^4}{6}(1+2\alpha+\beta)y^{(4)}
+O(h^5).
$$
The $hy'$ terms agree for every $\alpha,\beta$, so the method is always consistent and is generically of order one.
The method is explicit exactly when $1+\alpha=0$. With $\alpha=-1$, its $h^2$ defect vanishes only for $\beta=3/2$, so a generic explicit member has order one; that exceptional explicit member has order two and cannot satisfy the next order condition.
For the unrestricted method to have order at least three, the $h^2$ and $h^3$ coefficients must agree:
$$
1+2\alpha+\beta=\frac12,
\qquad
\frac{1-\beta}{2}=\frac16.
$$
Solving gives
$$
\boxed{\alpha=-\frac7{12},
\qquad
\beta=\frac23}.
$$
For these values, $1+2\alpha+\beta=1/2$, so the right-side $h^4$ coefficient is $1/12$, different from the left-side coefficient $1/24$. Hence the order is exactly three.
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