The first problem is that the author is considering lower case and upper case to be the same. So you should switch $52$ for $26$ in all your arguments
For the first question you are assuming that the two $T$'s are the last two letters, while this is not necessarily the case, to fix this first select the two positions that will have the $T$ (There are $6$ ways to do this). And after that use your argument, only you need to modify it since there are in fact $25$ options for each of the non-$T$ letters (Since $T$ is not allowed). So the answer should be $6\cdot25^2\cdot10^2\cdot1=375000$ as desired.
For the second question there are in fact $26^4\cdot10^3=456976000$ total Plates without any restrictions (I just used your argument switching $52$ for $26$).
Instead of counting the Plates that have at least one $T$ and at least on $4$ we shall count those that are lacking a $4$ or a $T$, and then substract this number from the total.
How many don't have a $4$ or don't have a $T$? We count those that don't have a $4$ and we count those that don't have a $T$ and add these numbers, and we subtract those that have no $4$ and no $T$ because those sequences were counted two times.(This idea is sometimes called inclusion-exclusion).
How many have no $T$? If they don't have $T$ there are only $25$ options for each letter, so there are $25^4\cdot10^3=390625000$
How many have no $4$? If they don't have $4$ there are only $9$ options for each digit, so there are $26^4\cdot 9^3=333135504$
How many don't have $T$ and don't have a $4$? There are $25$ options for each letter and $9$ options for each digit so $25^4\cdot9^3=284765625$.
So by our previous argument there are $390625000+333135504-284765626=438994879$ Plates that have no $4$ or no $T$. And so we have $456976000-438994879=17981121$ that have at least one $T$ and at least one $4$.
You are correct for the first problem, though I would rather think of $8!$ as a single high-level arranging operation rather than eight separate placements of the characters. I view this enumeration as consisting of three steps.
Step 1: Choose three distinct numbers in $\binom{10}{3}$ ways
Step 2: Choose five distinct letters in $\binom{26}{5}$ ways
Step 3: Arrange these eight characters in $8!$ ways
You can approach the second problem similarly if you think of the digits as comprising a single, indivisible unit (though the digits may be mixed up among themselves).
- Step 1: Choose three distinct numbers in $\binom{10}{3}$ ways
- Step 2: Choose five distinct letters in $\binom{26}{5}$ ways
- Step 3: Arrange the digits among themselves in $3!$ ways
- Step 4: Arrange the six items (five letters and one block of digits) in $6!$ ways
Best Answer
If no repetitions of letters then we have $\dbinom{26}{5}$ ways.
If no repetitions of digits then we have $\dbinom{10}{3}$ ways.
If no repetitions of letters or digits are allowed, then we have $$\dbinom{10}{3}\dbinom{26}{5}\cdot8!\mbox{ ways}$$