The work done by Tension and Normal is not always zero

In general if the situations are more complex then we have conservation laws and work energy theorem in our pocket.

Both of the situations are not much complex,in $1^{st}$ case:

Working from ground frame we can use work energy theorem:

$$Delta K=W_{tension}+ W_{gravity}$$

Here the problem shortens easily,because finding the work done by gravity is very easy as it is nothing but negative of change in potential energy of the bob-earth system and change in kinetic energy is already known to us by kinematics.

So the work done by gravity will come out to be equal to: $$W_{gravity}=-mgl(1-costheta)$$ Here $l$ is the length of rope and $theta$ is a variable angle.

For getting the final speed in $x$ i.e horizontal direction, just resolve the net centripetal acceleration in this direction,then the net acceleration $a_{x}=frac{dv_{x}}{dt}$ in horizontal direction is given as:

$$frac{dv_{x}}{dt}=a+frac{v_{b,c}^{2}sintheta}{l}$$

Similarly resolve the centripetal acceleration in $y$ direction,then

$$frac{dv_{y}}{dt}=frac{v_{b,c}^{2}costheta}{l}-g$$

Here, $$vec{v_{b,g}}=v_{x}hat x+v_{y}hat y$$ Just solve the above equations and after getting the final speed i.e $v_{b,g}=sqrt{v_{x}^2+v_{y}^2}$we can get the change in Kinetic energy.

Now work done by tension can be calculated easily

Using

$$dvec{s} = dvec{s_{b/c}} + dvec{s_{c/g}}$$ where, $dvec{s}$ is the small displacement of the body wrt ground.
$dvec{s_{b/c}}$ is small displacement of the mass wrt vehicle.
$dvec{s_{c/g}}$ is small the displacement of vehicle wrt to the ground.

we can write the work done by such forces as $$W_{F} = int_{A}^{B}vec{F}.dvec{s} = int_{A}^{B}vec{F}.(dvec{s_{b/c}} + dvec{s_{c/g}}) = int_{A}^{B}vec{F}.dvec{s_{b/c}}+int_{A}^{B}vec{F}.dvec{s_{c/g}}$$ But the such forces relative to the vehicle do not do any work. Hence, the first term vanishes
$$W_{F} = int_{A}^{B}vec{F}.dvec{s_{c/g}}$$

For illustration:
Let's have a look at the problem $1$ stated in the question. We need to find the work done on the mass by tension force in the journey from $A$ to $B$.
Now, the small displacement of the block $dvec{s}$ can be resolved in two parts, first due to motion of car (say $dvec{s_c}$) and second due to the circular motion of the mass around the pivot (say $dvec{r}$). The expression of work done by tension, becomes $$W_{T}=int_{A}^{B}vec{T}.dvec{s} = int_{A}^{B}vec{T}.vec{ds_c}+int_{A}^{B}vec{T}.vec{dr}$$
But, Tension is always perpendicular to $vec{dr}$. Hence, $$W_{T} = int_{A}^{B}vec{T}.dvec{s_c}$$