= Solution
Use the <Radon-Nikodym derivative> at $T_1$ and the <discounted bond price martingale> from part (b):
$$
\mathbb E^{\mathbb Q_{T_1}}P(T_1,T_2)
=\frac{\mathbb E^{\mathbb Q}[D_{T_1}P(T_1,T_2)]}{P(0,T_1)}
=\frac{P(0,T_2)}{P(0,T_1)}.
$$
For the <variance>, put $b_t^i=\int_t^{T_i}\sigma(t,u)du$ and $d_t=b_t^2-b_t^1=\int_{T_1}^{T_2}\sigma(t,u)du$, for $t\leq T_1$. The ratio $R_t=P(t,T_2)/P(t,T_1)$ has, by the <Itô formula>,
$$
d\log R_t=-\frac12\big((b_t^2)^2-(b_t^1)^2\big)dt-d_t\,dW_t.
$$
Under the <T-forward measure> $\mathbb Q_{T_1}$, $dW_t=dW_t^{T_1}-b_t^1dt$, so its drift becomes $-d_t^2/2$. Since $R_{T_1}=P(T_1,T_2)$,
$$
\log P(T_1,T_2)=\log\frac{P(0,T_2)}{P(0,T_1)}
-\frac12\int_0^{T_1}d_t^2dt-\int_0^{T_1}d_t\,dW_t^{T_1}.
$$
The first two terms are deterministic; the <Itô isometry> supplies the <variance> of the last. \b[The two requested answers are]
$$
\boxed{\mathbb E^{\mathbb Q_{T_1}}P(T_1,T_2)=\frac{P(0,T_2)}{P(0,T_1)},\qquad
\operatorname{Var}^{\mathbb Q_{T_1}}\log P(T_1,T_2)
=\int_0^{T_1}\left(\int_{T_1}^{T_2}\sigma(t,u)\,du\right)^2dt.}
$$
In fact the terminal <zero-coupon bond> price has a <log-normal distribution> under this <forward measure>, with the deterministic negative half-variance correction ensuring the displayed mean.
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