= Solution
For $h\lambda/2<1$, the implicit step map $z\mapsto y_n+(h/2)[f(t_n,y_n)+f(t_{n+1},z)]$ is a <contraction>, so the <Banach fixed-point theorem> supplies a unique next value. Smoothness of the exact solution gives the one-step defect
$$
d_n=y(t_{n+1})-y(t_n)-\frac h2\bigl[y'(t_n)+y'(t_{n+1})\bigr],\qquad |d_n|\leq Ch^3,
$$
uniformly on $[0,T]$, by the <trapezoidal rule> error estimate or a direct Taylor expansion. Let $e_n=y_n-y(t_n)$. Subtracting exact and numerical steps and applying the <Lipschitz condition> gives
$$
(1-h\lambda/2)|e_{n+1}|\leq(1+h\lambda/2)|e_n|+Ch^3.
$$
Put $r_h=(1+h\lambda/2)/(1-h\lambda/2)$. With $e_0=0$, iteration yields $|e_n|\leq[Ch^3/(1-h\lambda/2)](r_h^n-1)/(r_h-1)$. Since $r_h-1=h\lambda/(1-h\lambda/2)$, this is $Ch^2(r_h^n-1)/\lambda$. For sufficiently small $h$, $\log r_h\leq2\lambda h$, so $r_h^n\leq e^{2\lambda T}$ whenever $nh\leq T$. Therefore
$$
\boxed{\max_{nh\leq T}|e_n|\leq\frac C\lambda(e^{2\lambda T}-1)h^2\longrightarrow0.}
$$
This proves convergence with global order two directly from the defect bound and stability recurrence.
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