= Solution
With the normalization used here, the <Lasso> estimator minimizes
$$
\frac1{2n}\lVert Y-X\beta\rVert_2^2+\lambda\lVert\beta\rVert_1.
$$
Optimality at $\widehat\beta$ relative to the feasible point $\beta^0$ gives
$$
\frac1{2n}\lVert Y-X\widehat\beta\rVert_2^2+\lambda\lVert\widehat\beta\rVert_1
\leq\frac1{2n}\lVert Y-X\beta^0\rVert_2^2+\lambda\lVert\beta^0\rVert_1.
$$
The columns of $X$ are centered, so $X^T\mathbf1=0$ and the centered noise produces the same <score function> as $\varepsilon$. Expanding the two squared norms and cancelling the noise norm yields the standard <Basic inequality for the Lasso>
$$
\frac1{2n}\lVert X(\widehat\beta-\beta^0)\rVert_2^2
\leq\frac1n(\widehat\beta-\beta^0)^TX^T\varepsilon
+\lambda\lVert\beta^0\rVert_1-\lambda\lVert\widehat\beta\rVert_1.
$$
Thus the displayed inequality in the question has a factor-of-two typo: its left side should be $\lVert X(\widehat\beta-\beta^0)\rVert_2^2/(2n)$, or both terms on its right should be doubled. No scaling of the usual squared-error Lasso objective produces the three displayed coefficients simultaneously. Parts b and d explicitly ask us to use the stated inequality, so their requested constants follow from that stated version.
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