How to get named export node (more easily)

First of all, great project. Had fun using it!

Maybe someone knows a better way to do this, I found it a bit inconvenient for something I would expect to be easier.

Let's take this input file:

```ts
export { ExportA, ExportB } from './DepA';
export { ExportC } from './DepB';
export type { TypeA, TypeB } from './types';
```

Now I wanted to get the node of `ExportC` so I can find it's references in other source files:

```ts
const namedExports = sourceFile.getExportDeclarations().map(declaration => declaration.getNamedExports());
const exportSpecifier = namedExports.flat().find(namedExport => namedExport.getName() === 'ExportC');
const node = exportSpecifier?.getNameNode();
const references = node?.findReferencesAsNodes();
```

It works as intended, yet I think this is slightly tedious, maybe there's a better way?

Start by reproducing the TypeScript export example and reading the existing getExportDeclarations(), getNamedExports(), getNameNode(), and findReferencesAsNodes() APIs. Determine whether a simpler supported path can expose the named export's references; done means the proposed API is clear, implemented, and preserves the existing behavior for ExportA, ExportB, ExportC, and type exports.

Written by the indexing model from the issue text.