Using a juggling-check, you can test both null and undefined in one hit:

if (x == null) {

If you use a strict-check, it will only be true for values set to null and won't evaluate as true for undefined variables:

if (x === null) {

You can try this with various values using this example:

var a: number;
var b: number = null;

function check(x, name) {
    if (x == null) {
        console.log(name + ' == null');
    }

    if (x === null) {
        console.log(name + ' === null');
    }

    if (typeof x === 'undefined') {
        console.log(name + ' is undefined');
    }
}

check(a, 'a');
check(b, 'b');

Output

"a == null"

"a is undefined"

"b == null"

"b === null"

Yes. As of TypeScript 3.7 (released on November 5, 2019), this feature is supported and is called Optional Chaining:

At its core, optional chaining lets us write code where TypeScript can immediately stop running some expressions if we run into a null or undefined. The star of the show in optional chaining is the new ?. operator for optional property accesses.

Refer to the TypeScript 3.7 release notes for more details.

Prior to version 3.7, this was not supported in TypeScript, although it was requested as early as Issue #16 on the TypeScript repo (dating back to 2014).

As far as what to call this operator, there doesn't appear to be a consensus. In addition to "optional chaining" (which is also what it's called in JavaScript and Swift), there are a couple of other examples:

• CoffeeScript refers to it as the existential operator (specifically, the "accessor variant" of the existential operator):

The accessor variant of the existential operator ?. can be used to soak up null references in a chain of properties. Use it instead of the dot accessor . in cases where the base value may be null or undefined.

• C# calls this a null-conditional operator.

a null-conditional operator applies a member access, ?., or element access, ?[], operation to its operand only if that operand evaluates to non-null; otherwise, it returns null.

• Kotlin refers to it as the safe call operator.

There are probably lots of other examples, too.

Yes, as of Typescript 3.7 you can now do this via optional-chaining

person?.getName()?.firstName

gets transpiled to

let firstName = person === null || person === void 0 ? void 0 : (_person$getName = person.getName()) === null || _person$getName === void 0 ? void 0 : _person$getName.firstName;

Note the check for null. This will work as expected if for example person is defined as

let person:any = null; //no runtime TypeError when calling person?.getName()

However if person is defined as

let person:any = {};//Uncaught TypeError: person.getName is not a function

See also this similar stackoverflow question

It's called the "Non-null assertion operator" and it tells the compiler that x.getY() is not null.

It's a new typescript 2.0 feature and you can read about it in the what's new page, here's what it says:

A new ! post-fix expression operator may be used to assert that its operand is non-null and non-undefined in contexts where the type checker is unable to conclude that fact. Specifically, the operation x! produces a value of the type of x with null and undefined excluded. Similar to type assertions of the forms x and x as T, the ! non-null assertion operator is simply removed in the emitted JavaScript code.

// Compiled with --strictNullChecks
function validateEntity(e?: Entity) {
    // Throw exception if e is null or invalid entity
}

function processEntity(e?: Entity) {
    validateEntity(e);
    let s = e!.name;  // Assert that e is non-null and access name
}

Edit

There's an issue for documenting this feature: Document non-null assertion operator (!)