I was surprised to hear that if $j: Y \to X$ is a closed embedding, then $j_*: D^b(Y) \to D^b(X)$ is in general not faithful (note you shouldn't expect fullness, since $\operatorname{Ext}^1(O_p, O_p) \cong CT_p$) ! Although one can show that for sheaves $F,G$, we have $\operatorname{Hom}^1(F,G) \hookrightarrow \operatorname{Hom}^1(j_*F,j_*G)$.
Could someone give an example of nonfaithfulness (for a regular closed embedding)?
Here is a "proof" that might hint how to find such an example.
Suppose we have $F,G \in D^b(Y)$ and we want to show $\operatorname{Hom}(F,G) \hookrightarrow \operatorname{Hom}(j_*F,j_*G)$. It's enough to prove the theorem locally (via the base change formula). Locally suppose we have a bundle $V_X$ with $s \in H^0(V)$ cutting out $Y$, so that we have the koszul resolution $$\wedge^r V^* \to ... \to V^* \to O_X \cong j_*O_Y.$$
We are working locally I remind you. Take $\operatorname{Hom}_Y(F,G)$, first resolve $F$ using only copies of $O_Y$ (which is projective locally so the hom describes the actual hom in the derived category if we use the resolution) via the resolution of $F$ we denote $$I := ... \to I_1 \to I_0 \cong F.$$ We then consider $\operatorname{Hom}_X(j_*I,j_*G)$, now $j_*I$ is not projective anymore (since $j_* O_Y$ isn't) but we can resolve each $j_* O_Y$ using the Koszul resolution described above. Now I want to say (and this is the suspicious part) that we can resolve $I$ by pointwise resolving using the Koszul resolution (i.e if we have a map $f:O_Y \to O_Y$ we can lift $f$ to $\tilde{f}: O_X \to O_X$ and so on for the complex koszul complex). If the extension claim is true, then when we use $\operatorname{Hom}(*,F)$ all the koszul maps vanish becoming a direct sum the first summand of which is $O_X \otimes O_Y = O_Y$ which would imply faithfulness.
The extension claim is probably not true; while $O_X \to O_Y$ is surjective we won't be able to consistently lift stuff into a complex. For instance the horseshoe lemma forces one of the resolutions and doesn't let you control all of them.
Consider the Segre embedding $$ Y \cong \mathbb{P}^1 \times \mathbb{P}^1 \hookrightarrow \mathbb{P}^3 = X $$ and let $F_1 = \mathcal{O}(1,0)$, $F_2 = \mathcal{O}(-1,-2)$ be line bundles on $Y$. Then $$ \mathrm{Ext}^2(F_1,F_2) \cong H^2(Y, \mathcal{O}(-2,-2)) = \Bbbk. $$ On the other hand, the sheaf $j_*F_1$ has a simple locally free resolution $$ 0 \to \mathcal{O}(-1)^{\oplus 2} \to \mathcal{O}^{\oplus 2} \to j_*F_1 \to 0. $$ Now, we have $$ \mathrm{Ext}^\bullet(\mathcal{O},j_*F_2) \cong H^\bullet(Y, \mathcal{O}(-1,-2)) = 0, $$ $$ \mathrm{Ext}^\bullet(\mathcal{O}(-1),j_*F_2) \cong H^\bullet(Y, \mathcal{O}(0,-1)) = 0. $$ Using this and the above resolution it is easy to see that $$ \mathrm{Ext}^\bullet(j_*F_1,j_*F_2) = 0. $$ In particular, the morphism $$ j_* \colon \mathrm{Ext}^2(F_1,F_2) \to \mathrm{Ext}^2(j_*F_1,j_*F_2) $$ is not injective.